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gamingwings gamingwings
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13 years ago
Reaction of zinc metal and tincture of iodine.
Rewrite the overall reaction: a)Zn (s) + I_2 (alcohol) -> Zn^2+ 2I-
What is the oxidation half-reaction? b)_____________________________
What is the reduction half-reaction? c)_____________________________ _
What is the oxidation number of Zinc as:
Zn (metal)? d)0
Zn^2+ ? e)+2
I_2? f)0
I-? g)1
In this reaction, zinc is h)reduced.
In this reaction, iodine is i)oxidized.
What is the oxidizing agent? j)zinc
What is the reducing agent? k)iodine


Can you help me with the ones I did not do, and tell me if any of the ones I did was wrong and the correct answer?
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wrote...
#1
Educator
13 years ago Edited: 13 years ago, bio_man
Oxidation of Zinc: Zn(s) Rightwards Arrow Zn2+ (aq) + 2e-
 
Reduction of Iodine: I2 (s) + 2e- Rightwards Arrow 2 I-(aq)

Net Reaction: Zn(s) + I2 (s) Rightwards Arrow ZnI2 (s)

The oxidation number for zinc is 2, so e)+2

In this reaction, zinc is h) reduced. TRUE ... The next one is true too because iodine loses electrons. The next two are also true...

Good job.
Answer accepted by topic starter
padrepadre
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#2
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