I have to write ionic equations for the following and I am unsure of how to do so. Here is the question and my attempt at each:
a) chromium dipped into silver nitrate
Cr3+ +Ag+(NO3)-
Cr 3+(NO3)- +Ag+
b) gold immersed in hydrocholoric acid
NO REACTION and therefore NO EQUATION
c) nickel pellets dropped into a bath of calcium acetate
Ni2+ + Ca2+ (C2H3O2)2-
Ca2+ + Ni 2+ (C2H3O2)2-
d) aluminium dropped into a bath of sulphuric acid
Al3+ + H22+(SO4)2-
H22+ + Al3+(SO4)2-
e) zinc dipped inot lead(II) nitrate
Zn2+ + Pb2+ (NO3-)2
Pb 2+ +Zn 2+ (NO3)2
I have the rules for writing the equations but the examples given are not helping me answer these questions. I think my mistakes are:
1) I am only supposed to ionize (aq) substances and I ionized everything
2) when writing a compound I am supposed to just give the charge of the whole molecule instead of each part.
Other than that I am lost.
) yes, this is the problem. "Chromium" is Chromium metal, Cr NOT the ion, likewise Nickel is Ni, etc.
2) You can give the charge for each part, that's fine, and even write them seperately as this will make it easier to spot the spectator ions.
eg. Lithium dropped into aqueous HBr
HBr(aq)
---> H+(aq) + Br-(aq)
Li(s) + H+(aq) + Br-(aq) -----> LiBr(aq) + 1/2H2(g)
But,
LiBr(aq)
---> Li+(aq) + Br-(aq)
So overall,
Li(s) + H+(aq) + Br-(aq) -----> Li+(aq) + Br-(aq) + 1/2H2(g)
So Bromide is a spectator ion, and the net ionic equation is
Li(s) + H+(aq) -----> Li+(aq) + 1/2H2(g)
Okay so a) might look like this:
a) chromium dipped into silver nitrate
Cr(s) +Ag+(NO3)- (aq)
Cr3+(NO3)- +Ag(s)
Then seperate the aqueous ions right? so:
Cr(s) + Ag+(aq) + NO3-(aq)
Cr3+(aq) + NO3-(aq) + Ag(s)
Then we can see NO3 is a spectator ion so we would rewrite it as:
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
Cr(s) +Ag+(NO3)- (aq)
Cr3+(NO3)- +Ag(s
Your formula for Chromium(III) Nitrate is wrong and the charge is not balanced. It should be Cr(NO3)3
Quote
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
Your previous mistake carries forward itself here. It resulted in your charges are not balanced. It should be:
Cr(s) + 3 Ag+(aq)
Cr3+(aq) + Ag(s)
Redox reactions - in which charge is transferred between reagents - can be balanced by inspection, although it can be extremely difficult. But lets start with very simple example:
Cu + Fe3+ -> Cu2+ + Fe2+
At first sight equation seems to be already balanced, but if we check charges - it is not. There is +3 charge on the left and +5 charge on the right, so we have to do something about it. Using inspection method we should select most complicated particle first. Let's say it is Fe3+:
Cu + 1Fe3+ -> Cu2+ + Fe2+
If so, we can already add 1 in front of Fe2+:
Cu + 1Fe3+ -> Cu2+ + 1Fe2+
Trying to balance copper will lead us nowhere (atoms are balanced, but charges will be left unbalanced as they were from the beginning), so let's look at the charge - it can be balanced just like atoms. We have +3 on the left and +2 on the right - so we need additional +1 on the right side. Do you remember that we can use fractions at this stage? Half Cu2+ will do:
Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+
What is wrong now is the left side of equation - too much copper at the moment. Once again, half will do:
1/2Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+
Final touch - multiply everything by 2 to remove fractions:
Cu + 2Fe3+ -> Cu2+ + 2Fe2+
Atoms - balanced (one of copper and two of iron on both sides). Charge - balanced (+6 on both sides). We have just balanced redox equation using inspection method. But in the case of more difficult ones - like for example
FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O
this approach is a waste of time. Much easier and faster ways of balancing redox reactions are oxidation numbers method and half reaction method.
Oxidation takes place when electrons are removed from an atom or ion and reduction is the process by which an atom or ion gains electrons. Oxidation (loss of electrons) and reduction (gain of electrons) must always accompany one another in a reaction.
An increase in oxidation number is called oxidation; and a reduction in oxidation number is called reduction.
A substance that gains electrons in a reaction is called an oxidizing agent. A substance that transfers or loses electrons in a reaction is called a reducing agent. The stronger the tendency of an oxidizing agent to gain electrons, the greater its strength. The weaker the tendency for a reducing agent to hold electrons, the greater is its strength as a reducing agent. See the table of reduction potentials.
Data and Observations
1. Write the balanced equations for the reactions in Part 1 (only where a reaction occurred).
Cu+2 + Pb
if nothing happened then write N.R. for no rxn
if a rxn occurs we have to indicate the products. Since this is a redox equation, I know electrons are being transferred. I remember Pb has a +2 charge from nomenclature. Cu has a +1 or +2. How do I know which to use? Well, your lab tells you you start with +2 and the observation says that you have a copper colored metal coating the Pb strip. This, of course, is Cu 0
Cu+2 + Pb
Cu + Pb+2
The good news is all these have a +2 charge as ions and a 0 charge as an element so they equation will be balanced as written.
Your observations will tell you whether to write an equation or not.
Keep this with your observations. On the facing page WRITE THE CORRECT RESULTS AND EQUATIONS. I will give those to you now.
Cu+2 + Pb
Cu + Pb+2 You should have seen Cu "growing" on the lead piece
Cu+2 + Zn
Cu + Zn+2 You should have seen Cu "growing" on the Zinc piece
Pb+2 + Cu
NR
Pb+2 + Zn
Pb + Zn+2 You should have seen Zn (gray) "growing" on the lead piece
Zn+2 + Pb
NR
Zn+2 + Cu
NR
2.Which metal is the strongest reducing agent? Which metal is the weakest reducing agent?
To answer this we need to look and see which metal had the most rxns. (which test tubes showed products)
in the first two rxns above Cu+2 has been changed to Cu. The ox # has gone down = reduction so the Pb is the reducing agent. Same for the rest of the rxns. What we see is
the Pb can reduce Cu+2
Zn can reduce Cu+2
Zn can reduce Pb+2
Since Zn reduces the other two it is the strongest reducing agent. Cu can?t reduce anything so it is the weakest.
3. Use the reactions in Part 1 to prepare a table of half-reactions with the strongest oxidizing agent at the top of the table. (Example: Cu2+ + 2e- ® Cu.) See attached table for example
Use your results from Question #2. If Zn tubes had 2 rxns happen, and Cu had one then I know Zn is better at giving electrons away than Cu. Giving away electrons is oxidation. So with the example Zn would be a better at giving away electrons than Cu. You have to be able to think backwards too. If Zn is better at giving than Cu then Cu is better at accepting than Zn. Giving away makes it a reducing agent and taking makes it an oxidizing agent. So Cu is a stronger oxidizing agent than Zn
so I would see
Cu2+ + 2e-
Cu then
Pb+2 + 2e-
Pb then
Zn+2 + 2e-
Zn
4. The molecular equation for the reaction in Step 10 is given. Split all soluble substances into ions
Fe+2 + SO4-2 + K+ +MnO4- + H+ +SO4-2
Fe+3 + SO4-2 + Mn + SO4-2 + K+ + SO4-2 + H2O
Eliminate substances that are identical and on opposite sides of the arrow
Fe+2 +MnO4- + H+
Fe+3 + Mn + H2O this shows an acidic environment so do a normal reodx ½ rxn balance
To label the reducing agent and oxidizing agents remember they must be reactants. If electrons are on the reactant side of the balanced ½ rxn, this is reduction. A substance that is reduced is classified as an oxidizing agent. If electrons are on the right this is oxidation and the reactant Is classified as a reducing agent
Questions and Conclusions
1.Using the Standard Reduction Table, List the half-reactions for the halogens in order with the strongest oxidizing agent on top.
To do this you need to look at the table on the last page of your lab. Find the Halogens (Group VII). The table shows each substance gaining electrons (reduction) so this is called a reduction table. The Eo column gives us the relative values for each rxn. Looking at the table F2 (a halogen) has a value of +2.87, the largest value in the chart. This makes it the best at gaining electrons. (This makes a little sense since you know that Fluorine is the most electronegative element we have. Electronegativity is the tendency to gain electrons) The larger the value, the better it is at gaining electrons. (This is another way to say oxidizing agents because oxidizing agents have to accept the electrons they take from another substance)
2. Write the balanced total reaction for Zn + Cu+2, Zn + Pb2+, and Pb + Cu+2.
Same as the rxns we wrote for the observation
3. Using The Standard Reduction Table, note that a reducing agent (on the right side of the table) has the possibility of reacting with any oxidizing agent (on the left). (The rate of the reaction and the concentration of ions are not taken into account.)
After studying the table, name the substance which can:
a. reduce Pb+2 to Pb but cannot reduce Ni+2 to Ni
Look for something on the right that is above Ni+2 but below Pb+2
b. Oxidize I- to I2 but cannot oxidize Br-1 to Br2
This is reverse of the one above so look for something on the right above I- but below Br-
4. The following equations describe the reaction of copper with nitric acid. Using the first, show both the oxidation and the reduction half-reactions.
Cu + HNO3
Cu+2 + NO + H2O (NO is colorless)
Copper is going from Cu which is 0 to Cu+2 (figure out what this is)
So some other reactant must be doing the opposite. Eliminate the spectators in the beginning so HNO3 becomes NO3-1 (what does it change into?), Now just balance like our homework. (O?s by adding water, H+ to balance H?s, etc)
NO + O2
NO2 (NO2 is brown and noxious)
ACIDS and BASES
Acids = substances which ionize to form H+ in solution, usually referred to as the HYDRONIUM ION H3O+
Common acids are HCl, HNO3, CH3COOH (acetic acid or vinegar), lemon or lime juice (citric acid), vitamin C (ascorbic acid).
Bases = substances which react with the H+ ions formed by acids, usually by causing an increase in the OH-concentration in aqueous solution. The most common bases are NaOH, KOH, and Ca(OH)2.
another base which does not contain OH- is NH3 (ammonia).
when NH3(g) dissolves in water it forms NH4+ and OH-.
Ammonia is a weak base, and is a weak electrolyte.
Strong acids and bases...see lecture.
Classification of Electrolytes
Compounds can be classified as strong electrolyte, weak electrolyte, and non electrolyte by looking at their solubility.
If a compound is water soluble and ionic, then it is probably a strong electrolyte.
If a compound is water soluble and not ionic, and is a strong acid, then it is a strong electrolyte.
Similarly, if a compound is water soluble and not ionic, but is a strong base, then it is a strong electrolyte.
If a compound is water soluble and not ionic, and is a weak acid or weak base, then it is a weak electrolyte.
Otherwise, the compound is probably a non-electrolytes.
Neutralization and salt formation
Acid + base -----> salt + water
A salt is a compound in which the anion is from an acid and the cation is from a base.
IONIC EQUATIONS.
These are used to highlight reactions between ions
See lecture for details.
REACTIONS IN AQUEOUS SOLUTIONS.
Metathesis Reactions
Metathesis reactions involve swapping ions in solution:
AX + BY -----> AY + BX.
Metathesis reactions will lead to a change in solution if one of three things occurs:
1. an insoluble solid is formed (precipitate),
2. weak or nonelectrolytes are formed, or
3. an insoluble gas is formed.
1. Precipitation Reactions
A precipitate is an insoluble solid that forms when two solutions are mixed.
TO KNOW WHETHER SOMETHING IS SOLUBLE OR NOT...
A solute is soluble in water if more than 0.01 mol of the substance
will dissolve in enough water to make 1 liter of solution.
LEARN SOLUBILITY RULES P. 97!
Consider
2KI(aq) + Pb(NO3)2(aq) ----> PbI2(s) + 2KNO3(aq)
Both KI(aq) + Pb(NO3)2(aq) are colorless solutions.
When mixed, they form a bright yellow precipitate of PbI2 and a solution of KNO3.
The final product of the reaction contains solid PbI2, aqueous K+ and aqueous NO3- ions.
The molecular equation lists all the species as molecules:
2KI(aq) + Pb(NO3)2(aq) -----> PbI2(s) + 2KNO3(aq)
However, we know that certain substances exist as ions in solution.
The full ionic equation lists all ions:
2K+(aq) +2I-(aq) + Pb2+(aq) + 2NO3-(aq) -----> PbI2(s) + 2K+(aq) + 2NO3-(aq)
The net ionic equation cancels spectator ions that are unchanged:
2I-(aq) + Pb2+(aq) ----> PbI2(s)
2. weak or nonelectrolytes are formed
E.g. Dissolution of Mg(OH)2 in Acid
Milk of magnesia is a suspension of Mg(OH)2 in water.
Mg(OH)2 is relatively insoluble in neutral water.
In acidic solutions the Mg(OH)2 dissolves.
In acidic solution, the hydronium ions (H3O+) react with the OH- from the Mg(OH)2 to form water.
In the process Mg2+ ions are free to move about the solution.
If HCl is used as the acid, the overall chemical equation is
Mg(OH)2(s) + 2HCl(aq) -----> MgCl2(aq) + 2H2O(l)
The net ionic equation is
Mg(OH)2(s) + 2H3O+(aq) -----> Mg2+(aq) + 4H2O(l)
Or Iron (III) Oxide with nitric acid.....see lecture
3. if insoluble gases are formed
E.g. FeS + 2HCl ----> H2S + FeCl2
Net ionic equation is
FeS + 2H+ ----> H2S + Fe2+
Introduction to Oxidation-Reduction Reactions
Oxidation and Reduction
Oxidation = Loss of Electrons
e.g. Zn ------> Zn2+ + 2 e-
Here zinc has been oxidized to zinc (II) ions
Reduction = Gain of Electrons
e.g. 2H3O+ + 2 e- ------> H2 + 2H2O
Here protons (represented as the hydronium ion) have been reduced to hydrogen gas.
In all reduction-oxidation (redox) reactions, one species is reduced at the same time another is oxidized.
The species that causes oxidation is called the oxidizing agent.
The species that causes reduction is called the reducing agent.
The oxidizing agent is always reduced and the reducing agent oxidized.
The substance that is oxidized loses electrons to the substance that is reduced.
Oxidation of Metals be Acids and Salts
It is common for metals to produce hydrogen gas when they react with acids. e.g. Mg and HCl (aq)
Mg(s) + 2HCl(aq) ------> MgCl2(aq) + H2(g).
In the process the metal is oxidized and the H+ is reduced.
It is possible for metals to be oxidized in the presence of a salt:
Fe(s) + Ni(NO3)2(aq) ------> Fe(NO3)2(aq) + Ni(s).
The net ionic equation shows the redox chemistry well:
Fe(s) + Ni2+(aq) ------> Fe2+(aq) + Ni(s).
In this reaction iron has been oxidized to Fe2+ while the Ni2+ has been reduced to Ni.
The Activity Series
We can list metals in order of decreasing ease of oxidation.
This list is the activity series.
A metal in the activity series can only be oxidized by a metal ion below it.
The metals at the top of the activity series are called active metals.
The metals at the bottom of the activity series are called noble metals.
Formation of Silver Crystals on Copper Wire :(Demonstration : narrative)
Copper wire is placed in a beaker.
A solution of silver nitrate is added to the beaker.
Copper reduces Ag+ to Ag.
The Cu is oxidized to Cu2+
The full molecular equation
2AgNO3(aq) + Cu(s) ------> Cu(NO3)2(aq) + 2Ag(s)
The net ionic equation is
2Ag+(aq) + Cu(s) -----> Cu2+(aq) + 2Ag(s)
Note that the solution changes from colorless to green indicating the presence of copper(II) nitrate.
Silver crystals form on the Cu wire.
Solution Stoichiometry and Chemical Analysis
Recognize that there are two different types of units:
1. laboratory units (the macroscopic units which we measure in lab) and
2. chemical units (the microscopic units which relate to moles).
Always convert the laboratory units into chemical units first.
Grams are converted to moles using molar mass.
Volume or molarity are converted into moles using M = mol/L.
Use the stoichiometric coefficients from the balanced chemical equation to move between reactants and product.
Convert the laboratory units back into the required units.
Moles are converted to grams using molar mass.
Moles are converted to molarity or volume using M = mol/L.
Example worked in lecture.
Titrations
A titration is an experiment in which the molarity of a substance is measured by knowing the molarity of another substance.
Example: Suppose we know the molarity of an NaOH solution and we want to find the molarity of an HCl solution.
WE know....
molarity of NaOH, volume of HCl.
What do we want?
Molarity of HCl.
What do we do?
Take a known volume of the HCl solution (20.0 mL, say) and measure the number of mL of NaOH solution required to react completely with the HCl solution.
What do we get?
Volume of NaOH. Since we already have the molarity of the NaOH, we can calculate moles of NaOH.
Next step?
We also know HCl + NaOH -----> NaCl + H2O. Therefore, we know moles of HCl.
Can we finish?
Knowing mol (HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.
Example will be worked in lecture.
simple example: NH3 (aq) + ClO- (aq) ® N2H4 (l) + Cl- (aq)
The half-reactions will be : NH3 ® N2H4 and ClO- ® Cl-
So we get 2 NH3 ® N2H4 + 2H+ + 2e-
and 2H+ + 2e- + ClO- ® Cl- + H2O
Another example: SO32- + MnO4- ® SO42- + Mn+2 in acid solution
(sulfurous acid) permanganate (sulfate)
SO3-2 ® SO42- + e- (oxidation)
MnO4- + e- ® Mn+2 (reduction)
To balance first eq. need to add H+ and H2O:
SO3-2 + H2O ® SO42- + 2e- + 2 H+
Need H2O on RHS to balance O: MnO4- + e- ® Mn+2 + 4H2O
Then need H+ on LHS to balance charge and H's: MnO4- + 8H+ + 5e- ® Mn+2 + 4H2O
However, one half-reaction has 5e's the other 2, so 10 is the lowest common denominator:
5SO32- + 5H2O ® 5SO42- + 10e- + 10H+
and 2MnO4- + 16H+ + 10e- ® 2Mn+2 + 8H2O
So adding these two together we get: 2MnO4- + 5SO32- + 6H+ ® 2Mn+2 + 3H2O + 5SO42-
Check that the answer is balanced etc.
Oxidation of zinc by copper: A neat demonstration is to take a strip of zinc metal (its a shiny metal which looks like silver) and stick it in a solution of copper sulfate (which is a beautiful deep blue). After a while the solution becomes colorless and the zinc strip becomes copper-colored, because it has become coated with copper!
The reaction is Zn + CuSO4 (blue) ® Cu + ZnSO4(colorless)
The oxidation half-reaction is:
Zn ® Zn+2 + 2e-
The reduction half-reaction is:
Cu+2 + 2e- ® Cu
Since the number of e's in each half-reaction is the same we just add the two to get the final balanced equation:
Zn + CuSO4 ® Cu + ZnSO4
Suppose we set up the system so that we could transfer the e-'s from one half-reaction to theother, we could then use the system to generate electricity! This is the principle of what's known as an
ELECTROCHEMICAL CELL
and is the basis of many common batteries. If we just stuck a piece of Zn in, say, a solution of zinc sulfate, and a piece of copper in a solution of copper sulfate, and connected the metals by a wire with a voltmeter in the circuit, we would find that nothing happened! Why not? Because the set-up would separate charge in the sense of making more Zn2+ than sulfate in one compartment and less Cu2+ than sulfate in the other. So what we need is a way of transporting ions from one compartment to another. This is usually done with what is known as a salt-bridge or a porous divider.
(see figure).
The purpose of the salt bridge is to allow ions to transfer to maintain electrical balance in each electrode compartment. The cell then works as follows:
Electrons are lost by the zinc (oxidation): Zn ® Zn+2 + 2e-
which generates zinc ions - these go into the solution and are neutralized by sulfate (coming through the salt-bridge from the adjacent compartment). The electrons flow through the wire to the copper electrode (a conductor in the solution - usually either the metal undergoing oxidation or reduction, or a "neutral" conductor such as carbon or platinum). The voltmeter records the electrical potential, which is analogous to gravitational potential (a useful analogy is to consider electrons similar to water - both flow "downhill".)
For 1 M solutions of zinc and copper sulfate (standard conditions) the voltage produced by this cell is 1.1 V. A volt is a measure of the electrical potential, or if you like, the "electron pressure".
In the second part of the cell the copper ions become reduced (Cu2+ + 2e- ® Cu), thus copper ions are lost from the solution and deposit as copper metal on the electrode (it can be made of any conductor). The loss of the copper ions leads to an excess of sulfate which migrates across the salt-bridge to the other compartment. (The salt bridge need not be made of a sulfate salt.)
A few important definitions:
Oxidation occurs at the anode (remember O and A come before C and R) (and is marked negative - because its where the electrons are released).
Reduction occurs at the cathode
Anions flow to the anode
Cations flow to the cathode
An electrolyte is a salt solution that allows the flow of both a current (and potentially ions).
Electrode - a conducting material, may or may not be part of the oxidation or reduction half-reactions. Common examples are platinum, carbon, zinc.
ELECTRICAL POTENTIAL or EMF (electromotive force)
An electron moves from an electrode of higher electrical potential to one of lower potential, i. e. energetically downhill. In so doing the electron can do work. The amount of work done is proportional to the number of electrons (the charge) and the potential energy difference,
Electrical work = charge x potential difference
A Coulomb (C) is the quantity of charge that passes a point in an electrical circuit when a current of 1 amp flows for 1 sec.
The charge on one electron is 1.602 x 10-19 C.
Electrical potential is measured in volts where 1 V is defined as 1 J of work performed when 1 C of charge passes through a potential difference of 1 V. i. e. 1 V = 1 J/C
Since the potential of a cell depends on the concentration of reactants, electrode potentials are given as standard potentials for conditions of 1 M or 1 atm.
They are denoted as Eo and are normally for 25° C.
For half-reactions the Eo potentials are called standard reduction potentials.
They are measured using the hydrogen electrode cell (2 H+ (aqueous, 1M) + 2 e- ® H2 (g)(1 atm) - with a Pt electrode) as one half-reaction, with its standard potential chosen as 0 V.
Any electrode at which a reduction half-reaction shows a greater tendency to occur than does 2 H+ + 2 e- ® H2 has a positive Eo. (And vice versa)
By convention, standard electrode potentials are given for the half-cell reaction written as a reduction e. g.
Zn+2 + 2e- ® Zn Eo = -0.76 V
Cu2+ + 2e- ® Cu Eo = +0.337 V
Eocell = Eoox + Eored = +0.76 + 0.34 = 1.1 V (as we will see shortly, if the cell potential is positive it means the reaction is spontaneous.)
NOTE for the oxidation reaction Zn ® Zn2+ + 2 e- we have to change the sign, since we are now considering the reverse reaction i. e. oxidation.
There are tables of standard electrode potentials e. g. Table 17-1 and in the appendix of your text.
Knowing the standard electrode potentials for each half-cell reaction in a cell allows us to calculate the cell potential. The more easily an element is reduced, the higher the standard electrode potential.
In the analogy with water, the more positive the Eo the higher the "elevation" of the electron, i. e. the more potential energy it has.
Here's an example: we could replace the Cu in the Cu/Zn cell with nickel. The measured cell Eo at 25° C is +0.51 V. What is the Eo for the Ni half-cell reaction?
We can tell that the Ni2+ gets reduced, because we see Ni plating out on the Ni electrode.
First we write the half-reactions:
Zn ® Zn2+ + 2e- Eo = +0.76 V
Ni2+ + 2 e- ® Ni Eo = ?
Net
Reaction Zn + Ni 2+ ® Zn2+ + Ni Eonet = +0.51 V (remember Eocell will always be +)
Therefore Eo for Ni2+ + 2 e- ® Ni must be 0.51 - 0.76 = -0.25 V.
Remember in Tables of standard reduction potentials:
The Eo values are for the oxidized form + electrons ® reduced form
If writing the half reaction in the reduced form ® oxidized form + electrons you must change the sign of Eo
All half-reactions are reversible
The more positive the value for Eo the better the oxidizing ability of the species on the left side of the reaction - thus F2 + 2e- ® 2 F- with Eo = +2.8 V is the best oxidizing agent in table 17-1, and the Li+ ion is the weakest oxidizing agent.
The more negative Eo is the more likely the species will occur in the opposite direction, i. e. Li metal is a very strong reducing agent.
The reaction of any species on the left in the Table with one below it on the right, will lead to a product-forming reaction i. e. a cell which works.
Electrochemical potentials depend on the nature of the reactants and products and their concentrations, NOT on the quantities of material used. I. e. the coefficients used to multiply half-reactions to balance the number of electrons do not affect Eo. For example
Fe3+ + e- ® Fe2+ Eo = +0.77 V
and 2 Fe3+ + 2 e- ® 2 Fe2+ Eo = +0.77 V
Electrolytes and Non-electrolytes
In solid NaCl the ions in the lattice are held in place by strong ionic bonds.
This means the ions cannot move about in an electric field.
Therefore, solid NaCl does not conduct electricity.
When NaCl is added to water, the salt dissociates into Na+(aq) and Cl-(aq).
In the presence of an electric field, the solution can conduct electricity.
Strong electrolytes exist in solution entirely (or almost entirely) as ions.
Non-electrolytes (e.g. sugar) do not conduct electricity at all. (ie. there are no ions in solution
Strong and Weak Electrolytes
Water has a low conductivity so the bulb does not burn in pure water.
Hydrogen chloride is soluble in water.
In water, HCl ionizes into H+ and Cl-.
Since HCl is a strong electrolyte, in water there are no HCl molecules, only H+ and Cl- ions.
When the HCl(g) is bubbled through the solution, the bulb glows brightly because of the presence of the ions.
When acetic acid replaces the HCl solution the bulb glows less brightly.
Acetic acid exists as a mixture of acetic acid molecules and H+ and acetate ions in solution.
Because acetic acid is a mixture of ions and parent molecules in solution, acetic acid is a weak electrolyte.
Therefore, the acetic acid solution does not cause the bulb to glow brightly.
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Dissolution of an Ionic Compound
Fig. 4.3 pg. 124
Properties of Solutes in Aqueous Solution
Ionic Compounds in Water
When an ionic compound dissolves in water, the ions dissociate.
This means that in solution, the solid no longer exists as a well ordered arrangement of ions in contact with each other.
Instead, each ion is surrounded by water molecules.
The positive ions have the oxygen atoms of water pointing towards the ion, negative ions have the hydrogen atoms of water pointing towards the ion.
The transport of ions through the solution causes electric current to flow through the solution.
When a molecular compound dissolves in water (e.g., CH3OH), there are no ions formed. Therefore, there is nothing in the solution to transport electric charge and the solution does not conduct electricity.
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Dissolution of NaCl
http://hogan.chem.lsu.edu/matter/chap26/animate1/an26_004.mov Dissolution of NaCl
When an ionic solid is placed in water, the ions dissociate.
If the attractions between the ions and water molecules overcome the ionic attractions in the lattice, the salt is soluble.
Cations are attracted to the O atoms in water (through lone pairs).
Anions are attracted to the H atoms in water.
In solution the ions are separated.
Potassium Iodide and Lead Nitrate
Fig. 4.4 pg. 126
Metathesis Reactions
Metathesis reactions involve swapping ions in solution:
AX + BY AY + BX.
Ion swapping will lead to a chemical reaction in solution if one of three things occurs:
an insoluble solid is formed (precipitate),
weak or nonelectrolytes are formed, or
an insoluble gas is formed.
Precipitation Reactions
A solute is soluble in water if more than 0.01 mol of the substance will dissolve in enough water to make 1 liter of solution.
A precipitate is an insoluble solid that forms when two solutions are mixed.
Consider 2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq)
Both KI(aq) + Pb(NO3)2(aq) are colorless solutions. When mixed, they form a bright yellow precipitate of PbI2 and a solution of KNO3.
The final product of the reaction contains solid PbI2, aqueous K+ and aqueous NO3- ions.
The molecular equation lists all the species as molecules:
2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq)
However, we know that certain substances exist as ions in solution.
The full ionic equation lists all ions:
2K+(aq) +2I-(aq) + Pb2+(aq) + 2NO3-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq)
The net ionic equation cancels those ions that are unchanged:
2I-(aq) + Pb2+(aq) PbI2(s)
Solubility of Ionic Compounds
Table 4.1 pg. 127
Metathesis Reactions
In order to determine whether or not a substance will dissolve or form a precipitate, certain rules apply.
These rules often have important exceptions.
Net Ionic Equations
Text slide.
Net Ionic Equations
Write down balanced chemical reaction described in problem.
Rewrite this reaction in ionic form (ionic equation).
Cancel out equal quantities of all ions which are identical on both sides of reaction arrow.
Convert the reactions below into net ionic equations:
Reaction between Na2CO3(aq) and MgSO4(aq)
Reaction between Pb(NO3)2 (aq) and Na2S(aq)
Reaction between (NH4)3PO4(aq) and CaCl2(aq)
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Space Filling Models of Acids
Top of Fig. 4.6 pg. 131
Properties of Solutes in Aqueous Solution
Acids
Definitions:
Dissociation = pre-formed ions in a solid move apart in solution.
Ionization = neutral substance forms ions in solution.
Acid = substances which ionizes to form H+ in solution.
Common acids are HCl, HNO3, CH3CO2H (acetic acid or vinegar), lemon, lime, vitamin C.
Shown here are HCl (top), nitric acid (middle), and acetic acid (bottom). O atoms are shown in red, H in white, Cl in green, N in blue and C in black.
Bases = substances which react with the H+ ions formed by acids.
Common bases are NH3 (ammonia), drano, milk of magnesia.
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Aqueous Acids
http://hogan.chem.lsu.edu/matter/chap26/animate3/an26_037.mov Introduction to Aqueous Acids
Acids increase the concentration of [H+] when dissolved in water.
When HCl is added to water it completely dissociates into H+ and Cl-.
HCl is a strong electrolyte.
There is no undissociated HCl at the end of the reaction, so HCl is a strong acid.
HNO3 is also a strong acid. Therefore, in water it forms H+ and NO3- with no undissociated HNO3 left.
Acetic acid is a weak acid. Therefore, it exists as a mixture of undissociated acetic acid, H+, and acetate ions in solution.
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Reaction Between Ammonia and Water
Bottom of Fig. 4.6 pg. 131
Properties of Solutes in Aqueous Solution
Strong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
Therefore, they are completely ionized in solution.
We write this as HCl H+ + Cl-.
Weak acids and bases are weak electrolytes.
Therefore, they are partially ionized in solution.
Since H+ is a naked proton, we refer to acids as proton donors and bases as proton acceptors.
In this figure, we see the proton transfer between NH3 (a weak base) and water (a weak acid). Since there is a mixture of NH3, H2O, NH4+, and OH- in solution, we write:
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Aqueous Bases
http://hogan.chem.lsu.edu/matter/chap26/animate3/an26_038.mov Introduction to Aqueous Bases
Bases increase the [OH-] concentration in aqueous solution.
The most common bases are NaOH, KOH, and Ca(OH)2.
When solid NaOH is added to water it completely dissociates into Na+ and OH- in solution.
Since there is no undissociated NaOH at the end of the reaction, NaOH is a strong base.
Strong bases are strong electrolytes.
NH3 is an example of a base which does not contain OH-.
However, when NH3(g) dissolves in water it forms NH4+ and OH-.
Ammonia is a weak base, so there is an equilibrium mixture of NH3, NH4+, and OH- in solution.
Most of the ammonia exists as NH3 in solution, it is a weak electrolyte.
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Strong Acids and Bases
Table 4.2 pg. 132
Strong Acids and Bases
Table 4.2 Common Strong Acids and Bases
Strong Acids Strong Bases
Hydrochloric, HCl Group 1A metal hydroxides (LiOH,
Hydrobromic, HBr NaOH, KOH, RbOH, CsOH)
Hydroiodic, HI .
Chloric, HClO3 Heavy group 2A metal hydroxides
Perchloric, HClO4 (Ca(OH)2, Sr(OH)2, Ba(OH)2)
Nitric, HNO3 .
Sulfuric, H2SO4 .
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Weak and Strong Electrolytes
Text slide.
Properties of Solutes in Aqueous Solution
Compounds can be classified as strong electrolytes, weak electrolytes, and nonelectrolytes by looking at their solubilities and classifications.
Water-soluble and ionic = strong electrolyte (NaCl: ionic salt, NH4Cl: ionic weak acid, NaC2H3O2: ionic weak base).
Water-soluble strong acid or strong base = strong electrolyte (HCl: ionic strong acid, NaOH: ionic strong base).
Water-soluble but not ionic weak acid or base = weak electrolyte (NH3: molecular weak base, HC2H3O2: molecular weak acid).
Not water-soluble but ionic = weak electrolyte (CaSO4: "insoluble" ionic).
Water-soluble but not ionic = nonelectrolyte (CH3OH: soluble molecular).
Not water-soluble and not ionic: = nonelectrolyte (C6H6: insoluble molecular).
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Natural Indicators
http://hogan.chem.lsu.edu/matter/chap27/demos/dm27_010.mov Natural Indicators
An indicator is a substance that changes color as a function of pH.
Most indicators are dyes.
Some indicators can be extracted from natural products.
Example: the dye which is characteristic of the red color in red cabbage leaves is an indicator.
The colorless dye is extracted with methanol.
The dye turns yellow/green in base and red in acid.
Tea is another natural indicator. It changes from brown to yellow/orange in acid. In this experiment, lemon juice is used as the acid.
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Dissolution of Mg(OH)2 by Acid
Fig. 4.8 pg. 135.
Dissolution of Mg(OH)2 with HCl
Phillip's Milk of Magnesia bottle in Fig. 4.8 a contains milky mixture of Mg(OH)2 and water.
Concentrated HCl solution is added from clear bottle in Fig. 4.8 b.
All Mg(OH)2 has dissolved in Fig. 4.8 c.
Mg(OH)2(s) + 2 HCl(aq) MgCl2 (aq) + 2 HOH(l)
Net ionic equation: Mg(OH)2(s) + 2 H+(aq) Mg2+(aq) + 2 HOH(l)
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Magnesium Hydroxide
http://hogan.chem.lsu.edu/matter/chap26/animate3/an26_041.mov Dissolution of Mg(OH)2 in Acid
Milk of magnesia is a suspension of Mg(OH)2 in water.
Mg(OH)2 is relatively insoluble in neutral water.
In acidic solutions the Mg(OH)2 dissolves.
The solid Mg(OH)2 consists of layers of Mg2+ ions sandwiched between OH- ions.
In water there is an equilibrium between the solid Mg(OH)2 and the ions.
In acidic solution, the hydronium ions (H3O+) react with the OH- from the Mg(OH)2 to form water.
In the process Mg2+ ions are free to move about the solution.
If HCl is used as the acid, the overall chemical equation is:
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
The net ionic equation is:
Mg(OH)2(s) + 2H3O+(aq) Mg2+(aq) + 4H2O(l)
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Electron Transfer Reactions
Right margin pg. 139
Introduction to Oxidation-Reduction Reactions
Oxidation and Reduction
In all reduction-oxidation (redox) reactions, one species is reduced at the same time another is oxidized.
The species that causes oxidation is called the oxidizing agent.
The species that causes reduction is called the reducing agent.
The oxidizing agent is always reduced and the reducing agent oxidized.
The substance that is oxidized loses electrons to the substance that is reduced.
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Redox I
http://hogan.chem.lsu.edu/matter/chap26/animate1/an26_007.mov Oxidation-Reduction Reactions I
Consider zinc metal in the presence of oxygen at high temperature.
The zinc reacts to form ZnO.
Electrons are transferred from zinc to oxygen.
So, the zinc becomes Zn2+ while the oxygen becomes O2-.
Zinc has been oxidized (i.e. its oxidation state has increased) by oxygen (the species causing the oxidation).
Oxygen has been reduced (i.e. its oxidation state has decreased) by zinc (the reducing agent).
In general, the loss of electrons is oxidation: Zn Zn2+ + 2e-.
The gain of electrons is reduction: O + 2e- O2-.
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Redox II
http://hogan.chem.lsu.edu/matter/chap26/animate1/an26_008.mov Oxidation-Reduction Reactions II
When Zn is added to dilute acid, H2(g) and Zn2+ ions are formed:
Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
H3O+ attacks the zinc atoms and an electron is transferred from the hydronium ion (H3O+) to the zinc metal.
After two such transfers, zinc is oxidized to Zn2+ while H+ is reduced to hydrogen gas.
When zinc is placed in aqueous copper sulfate, a similar reaction occurs: Zn is oxidized by Cu2+ to Zn2+ while Cu2+ is reduced to Cu0 by Zn.
When zinc is oxidized it causes reduction.
Therefore, zinc is the reducing agent.
The substances that are reduced, cause oxidation and are the oxidizing agents.
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Oxidation of Mg by Acid
Fig. 4.13 pg. 141
Introduction to Oxidation-Reduction Reactions
Oxidation of Metals be Acids and Salts
It is common for metal to produce hydrogen gas when they react with acids. Shown here is the reaction between Mg and HCl:
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g).
In the process the metal is oxidized and the H+ is reduced.
It is possible for metals to be oxidized in the presence of a salt:
Fe(s) + Ni(NO3)2(aq) Fe(NO3)2(aq) + Ni(s).
The net ionic equation shows the redox chemistry well:
Fe(s) + Ni2+(aq) Fe2+(aq) + Ni(s).
In this reaction iron has been oxidized to Fe2+ while the Ni2+ has been reduced to Ni.
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Formation of Silver Crystals
http://hogan.chem.lsu.edu/matter/chap27/demos/dm27_006.mov Formation of Silver Crystals on Copper Wire
See Fig. 4.15 pg. 125
Copper wire is placed in a beaker (or test tube).
A solution of silver nitrate is added to the beaker (test tube).
Copper reduces Ag+ to Ag.
The Cu is oxidized to Cu2+.
The full molecular equation:
2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s)
The net ionic equation is:
2Ag+(aq) + Cu(s) Cu2+(aq) + 2Ag(s)
Note that the solution changes from colorless to blue indicating the presence of copper(II) nitrate.
Silver crystals form on the Cu wire.
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Solution Formation From Solid
Fig. 4.16 pg. 146
Solution Composition
Molarity
A solution is made when one substance (the solute) is dissolved in another (the solvent).
The solute is the substance that is present in smallest amount.
Solutions in which water is the solvent are called aqueous solutions.
Solutions can be prepared with different concentrations by adding different amounts of solute to solvent.
The amount (moles) of solution per liter of solution is the molarity of the solution.
By knowing the molarity of a quantity in liters of solution, we can easily calculate the number of moles (and, by using molar mass, the mass) of solute.
In this figure, blue copper(II) sulfate pentahydrate, CuSO4·5H2O, is weighed (62.4 g, 0.250 mol) and placed in a 250 mL volumetric flask. A little water is added and the flask swirled to ensure the copper sulfate dissolves. When all the copper sulfate has dissolved, the flask is filled to the mark with water. The molarity of the solution is 0.250 mol / 0.250 L = 1.00 M.
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Solution Formation
http://hogan.chem.lsu.edu/matter/chap26/animate1/an26_012.mov Solution Formation from a Solid
To form a solution of known concentration (standard solution) a certain amount of solid must first be accurately weighed.
To prepare 250 mL of 1.00 M CuSO4 solution, 0.250 moles of copper sulfate are required.
Copper sulfate occurs as the pentahydrate, CuSO4·5H2O (FW = 249.7 g/mol). Therefore, 62.40 g of CuSO4·5H2O are required.
The copper sulfate is added to a 250 mL volumetric flask.
Some water is added and the flask is swirled to dissolve the salt.
Once all the salt is dissolved, enough water is added to bring the solution to the mark on the volumetric flask.
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Dilution
http://hogan.chem.lsu.edu/matter/chap26/animate1/an26_011.mov Solution Formation By Dilution
See Fig. 4.19 pg.131
Suppose we want to prepare 250 mL of a 0.100 M solution of copper sulfate starting with a 1.00 M stock solution.
We carefully pipet 25.00 mL of the stock solution and add it to a 250 mL volumetric flask.
Some water is added to the volumetric flask and the flask is swirled to ensure good mixing.
Then enough water is added to make the total volume of the solution 250 mL.
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Volumetric Calculations
Text slide.
Volumetric Calculations
To work problems involving molarity, volume, and grams first calculate number of moles of solute you are working with.
(vol of solution) x (concentration) = (moles of solute)
(grams of solute) _ (MM of solute) = (moles of solute)
(vol of solute) x (density of solute) _ (MM of solute) = (moles of solute)
Next use dimensional analysis to calculate the final quantity you need.
How many g of K2Cr2O7 in 50.0 mL of 0.850 M solution?
What is molarity of 2.50 g of (NH4)2SO4 in a 250 mL solution?
How many mL of 0.387 M CuSO4 contains 1.00 g of solute?
How many g of AgNO3 needed for 100.0 mL of 0.200 M solution?
How much 6.0 M HNO3 needed to create 250 mL of 1.0 M HNO3 ?
Calculate molarity of 250 mL glycerol (C3H8O3) solution made using 50.000 mL of pure glycerol. Density = 1.2656 g/mL.
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Titration
Fig. 4.21 pg. 153
Solution Stoichiometry and Chemical Analysis
Titrations
A titration is an experiment in which the molarity of a substance is measured by knowing the molarity of another substance.
Example: Suppose we know the molarity of an NaOH solution and we want to find the molarity of an HCl solution.
First let us examine what we know: molarity of NaOH, volume of HCl.
What do we want? Molarity of HCl.
What do we do? Take a known volume of the HCl solution (20.0 mL, say) and measure the number of mL of NaOH solution required to react completely with the HCl solution.
What do we get? Volume of NaOH. Since we already have the molarity of the NaOH, we can calculate moles of NaOH.
Next step? We also know HCl + NaOH NaCl + H2O. Therefore, we know moles of HCl.
Can we finish? Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.
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Titration
http://hogan.chem.lsu.edu/matter/chap26/animate3/an26_042.mov Acid-Base Titration
A titration is an experiment in which the concentration of an unknown acid or base is deduced from the concentration of a standard solution of base or acid.
Consider the addition of a standard NaOH solution added to a solution of HCl.
A pH meter is used to measure the acidity of the solution.
When 0.100 M NaOH solution is added to a 0.100 M solution of HCl, the pH increases from 1.0.
As more base is added, the pH gradually increases.
Near the equivalence point (the point defined by stoichiometry as the point at which enough base has been added to neutralize the acid), the pH begins to change dramatically with small additions of base.
The last few drops of base change the pH from around 3 to 7.
At the equivalence point the pH is 7.0 because the solution is neutral.
Addition of one drop of NaOH after equivalence causes the pH to increase to about 10.0.
Note that the phenolphthalein indicator changes color from colorless to red after the equivalence point.
The end point of the titration is the point at which we observe a color change.
The end point is experimentally determined, the equivalence point is determined by stoichiometry
Claims:
We claim:
1. A method for preparing a chromate-treated zinc-plated steel strip, comprising
effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.
2. A method for preparing a chromate-treated zinc-plated steel strip, comprising
effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.
3. A method for preparing a chromate-treated zinc-plated steel strip, comprising
effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to substantially less than 30 C/dm.sup.2.
4. A method for preparing a chromate-treated zinc-plated steel strip, comprising
effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current quantity density of 1 to 50 A/dm.sup.2 and to an electric of 5 to substantially less than 30 C/dm.sup.2.
Description:
BACKGROUND OF THE INVENTION
This invention relates to chromate-treated zinc-plated steel strips having high corrosion resistance without coating, good coating adherence, and firm adhesive bond to vinyl chloride and similar resins, as well as a method for making the same.
Most of currently available zinc-plated steel strips are zinc electroplated steel strips and zinc hot dipped or galvanized steel strips. Since they are not necessarily sufficient in corrosion resistance, various zinc alloy plated steel strips including Zn-Ni, Zn-Fe, and Zn-Al alloy plated ones have been developed and marketed. These advanced products may be used as such, but are often used after a chromate treatment which serves for white rust prevention and as a primary treatment for subsequent coating.
Most currently used chromate treatments are reactive chromate treatments which are applied to those products which require a white rust generating time of 24 to 100 hours in the standard salt spray test. In the reactive chromate treatments, the quantity and nature of the resulting chromate film are largely affected by the reactivity of the underlying metal. More particularly, because of their relatively high reactivity, zinc-plated steel strips can be coated with a chromate film only by dipping the strips in conventional chromate solutions having a relatively low etching power. Since zinc alloy-plated steel strips, however, are low reactive, a chromate film can not fully grow thereon in the conventional chromate solutions. Although corrosion resistance is improved by increasing the quantity of a chromate film deposited, an excessively built-up chromate film turns to be yellow due to hexavalent chromium and thus exhibits an undesirable appearance. When such thickly chromated strips are coated with paint, the adherence between the chromate film and the paint is poor.
As a high speed plating line becomes widespread, post-treatment procedures also want speeding up. In order for the reactive chromate treatment to produce a competent quantity of a uniform chromate film, continuous dipping or spraying for a certain period of tim, typically 4 to 10 seconds is necessary. A common approach for accommodating with the high speed line is to increase the number of tanks to extend the reaction time.
Another class of chromate treatment including coating and electrolytic chromate treatments becomes recently available because these treatments are little affected by the reactivity of steel strips and take a short time to completion. The coating chromate treatment is applied to those products which require a corrosion resisting time of 200 hours or more in the standard salt spray test. The electrolytic chromate treatment results in more improved adherence to a coating as compared with the reactive and coating chromate treatments because the resulting chromate film consists essentially of trivalent chromium.
The coating chromate treatment is generally practiced by a method of adding colloidal silica as a film forming agent as disclosed in Japanese Patent Publication No. 42-14050. Another method for conducting the coating chromate treatment involves applying a chromate solution containing an organic polymer by roll coating or dipping and roll squeezing, followed by drying with or without water rinsing. The coating chromate treatment, however, has the disadvantages that it is difficult to control the quantity of a chromate film deposited and that a high speed treatment frequently invites inconsistencies because the chromate film tends to be nonuniform in a transverse direction to the feed direction. It is needed to develop a technique enabling uniform film formation. Another disadvantage is that the resulting chromate film has poor adherence to a coating because the film is thick and retains hexavalent chromium unchanged throughout the film. Also, the chromate film provides a poor adhesive bond to vinyl chloride and similar resins.
The electrolytic chromate treatment is applied by subjecting a steel strip to cathodic electrolysis whereby hexavalent chromium is electrically reduced to trivalent chromium to form a hydrated oxide film at the strip surface. The electrolytic chromate treatment can not only readily accommodate with speeding-up because the quantity of a chromate film can be controlled by a quantity of electricity, but also be applied to various types of steel strips because hexavalent chromium ions in the chromate solution are reduced electrically rather than by redox reaction. The chromate film resulting from the electrolytic chromate treatment consists essentially of trivalent chromium and has higher coating adherence as compared with the reactive and coating chromate treatments, but is less corrosion resistant as compared with the reactive chromate treatment.
One prior art method for carrying out an electrolytic chromate treatment is disclosed in Japanese Patent Publication No. 47-44417 which is incorporated herein by reference. This method is successful in forming a good, but thin chromate film only at a relatively low current density. The chromate layer cannot be further grown even by increasing electricity quantity. Differently stated, the method fails to form a thick chromate film on a zinc alloy plated steel strip. As previously indicated, in general, the electrolytic chromate film is less corrosion resistant as compared with the reactive and coating type chromate films having the same amount of chromium deposited. This is probably because the electrolytic chromate film tends to be porous due to evolution of hydrogen gas during film formation and because the chromate film composed mainly of trivalent chromium contains an insufficient amount of hexavalent chromium to seal such pores or defects, that is, lacks a self-healing ability.
Another method for carrying out an electrolytic chromate treatment is disclosed in Japanese Patent Application Kokai No. 60-110896 which is incorporated herein by reference. A chromate film is formed in a bath containing hexavalent chromium (Cr.sup.6+)+cationic colloidal silica+H.sub.2 SO.sub.4 +optional NaOH for pH adjustment. Due to the inclusion of sulfate residues in the bath, metallic Cr tends to deposit in a chromate film particularly at a high current density and thus, the chromate film often becomes black colored. The cationic colloidal silica and sulfate residues serve as film forming agents while processing inconsistencies often occur. An observation of chromate films under a scanning electron microscope has indicated that chromate films resulting from a bath containing a fluoride additive are more uniform and dense than those from a bath containing sulfuric acid.
SUMMARY OF THE INVENTION
Therefore, an object of the present invention is to provide a novel and improved chromate-treated zinc alloy-plated steel strip having a chromate film exhibiting high corrosion resistance, good adherence to a coating, and a firm adhesive bond to vinyl chloride and similar resins.
Another object of the present invention is to provide a method for making the same wherein an electrolytic chromate treatment can be carried out on any type of zinc alloy plating within a short time to a sufficient thickness of chromate film to meet the intended application.
In the initial of developing a zinc or zinc alloy electroplated steel strip having a chromate film exhibiting satisfactory corrosion resistance, coating adherence, and adhesive bond, we attempted to carry out a coating adherence improving treatment on a reactive chromate film. This attempt, however, requires two treatments. It also requires a choice between thick and thin films. A thick film must be formed to insure corrosion resistance when it is intended to use the final product without coating. A thin film will suffice when the final product is coated on use. A compromise is to form a chromate film of moderate thickness having a minimized content of hexavalent chromium in the outermost surface layer.
Intending to produce a chromate film fulfilling the requirements of corrosion resistance, coating adherence, and adhesive bond by only an electrolytic chromate treatment, we have discovered that the object can be attained by controlling the composition of a chromate film.
More particularly, it is desired that the outermost surface region or layer of a chromate film have an effective composition to provide corrosion resistance and coating adherence.
We have discovered it effective in enhancing corrosion resistance that (1) an appropriate amount of hexavalent chromium is contained in the chromate film predominantly comprising trivalent chromium to impart a self-sealing or self-healing ability, (2) the film thickness is increased to form a reinforced barrier by adding a film forming agent such as silicon dioxide, and (3) the film is rendered uniform by adding an etching agent.
We have also discovered it effective in enhancing coating adherence that (4) the outermost surface layer is a thin region composed predominantly of trivalent chromate. (5) SiO.sub.2 is effective in enhancing coating adherence, but tends to cause delamination in the chromate film as the film becomes thick. It will be advantageous that the chromate film be bonded to a resin laminated board with an adhesive. We have discovered that (6) the adhesive bond can be improved by adding Al.sub.2 O.sub.3 to the chromate bath along with SiO.sub.2. The present invention is predicated on these findings.
According to a first aspect of the present invention, there is provided a chromate-treated zinc-plated steel strip comprising
a steel substrate,
a zinc base plating on at least one surface of the substrate,
a metallic chromium layer on the zinc base plating,
a chromium oxide layer on the metallic chromium layer, consisting essentially of the oxide of trivalent chromium, and
an outermost surface layer on the chromium oxide layer, consisting essentially of silicon dioxide and oxides of a major proportion of trivalent chromium and an effective proportion of hexavalent chromium and hydrates thereof.
According to a second aspect of the present invention, there is provided a method for preparing a chromate-treated zinc-plated steel strip, comprising
effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.
In one preferred embodiment of the present invention, the outermost surface layer further contains aluminum oxide. In this case, the electrolytic chromate bath used in preparing a corresponding chromate-treated zinc-plated steel strip further contains 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol in addition to the above-defined ingredients.
BRIEF DESCRIPTION OF THE DRAWINGS
The above and other objects, features, and advantages of the present invention will be readily understood by reading the following description when taken in conjunction with the accompanying drawings, in which:
FIG. 1 is a diagram showing the proportions of metallic Cr, Cr.sup.3+, and Cr.sup.6+ in the chromate film analyzed by ESCA;
FIG. 2 is a diagram showing the relative proportions of Si and Cr in the chromate film analyzed by GDS;
FIG. 3 is a diagram showing the weight of chromium deposited as a function of electricity quantity in the chromate treatment of Example 1;
FIG. 4 is a diagram showing the percent white rust of chromate treated steel strips produced in Example 2 and Comparative Examples 2 and 3 as a function of salt spray test time;
FIGS. 5 and 6 graphically show the weight of chromium deposited as a function of electricity quantity in the chromate treatment of various zinc-plated steel strips in different baths in Example 3; and
FIG. 7 is a diagram showing the percent white rust of chromate treated steel strips produced in Example 4 and Comparative Example 4 as a function of salt spray test time.
DETAILED DESCRIPTION OF THE INVENTION
In the present disclosure, the term zinc plated steel strips is used to encompass steel strips plated with zinc and zinc based alloys. Typical examples of the zinc plated steel strips include zinc electroplated (or electrogalvanized), zinc hot dipped (or galvanized), galvannealed, Zn-Ni alloy plated, Zn-Fe alloy plated, and Zn-Al alloy plated steel strips. These plating surfaces are different in metal or alloy phase and particularly in reactivity during a subsequent treatment, for example, a heat treatment to form an oxide coating.
According to the present invention, electrolysis is effected on various zinc-plated steel strips in a chromate bath with the strips made cathode, by supplying constant current. Hexavalent chromium ions typically present in the form of Cr.sub.2 O.sub.7.sup.2- and CrO.sub.4.sup.2+ in the bath are electrochemically reduced to trivalent chromium ions to form a chromate film predominantly comprising Cr.sup.3 +. Thus the formation of chromate film is little affected by the underlying layer, that is, zinc plating. The amount of chromate film formed is proportional to a quantity of electricity supplied so that the thickness of chromate film may be controlled over a wide range from thin to thick films depending on the intended application of the chromated strip.
The chromate-treated, zinc-plated steel strip according to the present invention has a chromate film consisting of
(1) a layer most adjacent to the zinc base plating which consists of metallic chromium,
(2) an intermediate layer which consists essentially of the oxide of trivalent chromium, and
(3) an outermost surface layer which consists essentially of silicon dioxide (SiO.sub.2), optional aluminum oxide (Al.sub.2 O.sub.3), and oxides of a major proportion of trivalent chromium and an effective proportion of hexavalent chromium and hydrates thereof.
The metallic chromium layer disposed in direct contact with the zinc base plating is not critical in the practice of the present invention, but is naturally deposited in a small amount from the chromate bath operated under the electrolytic conditions according to the present invention. The metallic chromium layer may be discontinuous. Excess deposition of metallic chromium is undesirable because the amount of subsequently formed hydrated oxides is reduced. The weight of metallic chromium deposited is preferably limited to the maximum of 20 mg/m.sup.2
In the outermost surface layer, trivalent chromium and an effective proportion of hexavalent chromium coexist. The effective proportion of hexavalent chromium means a sufficient amount of hexavalent chromium to exert a full self-healing effect. The proportion of hexavalent chromium preferably ranges from 1/100 to 1/5 of the total weight of chromium in the chromate film. The lower limit of hexavalent chromium is set to 1/100 or 1% below which hexavalent chromium is too less to provide a self-healing effect, failing to improve corrosion resistance. The presence of hexavalent chromium in excess of 1/5 or 20% of the total weight of chromium will result in a colored film and detract from coating adherence.
FIG. 1 shows the proportions of metallic, trivalent and hexavalent chromiums based on the total weight of chromium in the chromate film according to the present invention. The proportions of Cr(0), Cr(III) and Cr(VI) are determined in a thickness direction of the film by electron spectroscopy for chemical analysis (ESCA) and expressed as their ratio to the total chromium.
The chromate film is preferably deposited to a weight of 20 to 200 mg/m.sup.2 calculated as Cr. A chromate film having less than 20 mg/m.sup.2 of Cr exhibits poor corrosion resistance without coating as demonstrated by the white rust generating time of about 24 hours in the standard salt spray test (SST). Conversely, a chromate film having more than 200 mg/m.sup.2 of Cr exhibits yellow color in appearance and poor coating adherence.
According to the present invention, the chromate film contains silicon dioxide (SiO.sub.2) The present invention is characterized in that silicon dioxide is preferentially present in the outermost surface layer. FIG. 2 illustrates the results of measurement of the chromate film by glow discharge spectrometry (GDS). The proportion of SiO.sub.2 preferably ranges from 1/40 to 1/2 of the total weight of chromium in the chromate film. Less than 1/40 of the total chromium weight of SiO.sub.2 is insufficient to exert its essential effect of film formation. The content of SiO.sub.2 is limited by such processing factors as transfer to rolls during manufacturing process. The presence of more than 1/2 of the total chromium weight of SiO.sub.2 results in a rather thick film and adversely affects coating adherence.
In one preferred embodiment of the present invention, the chromate film further contains aluminum oxide (Al.sub.2 O.sub.3) in its outermost surface layer. Aluminum oxide is introduced to enhance coating adherence and particularly, adhesive bond characteristics. The amount of Al.sub.2 O.sub.3 preferably ranges from 10 to 1/2 of the weight of SiO.sub.2. Inclusion of Al.sub.2 O.sub.3 in amounts of less than 1/10 of the SiO.sub.2
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