In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at th

Similar question:

In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at the cathode. The MnO2 used in dry cells can itself be produced by an electrochemical process of which one half-reaction is shown below.



If a current of 30.5 A is used, how many hours are needed to produce 1.15 kg of MnO2?

Answer:

Mn2+(aq) + 2 H2O(l) => MnO2(s) + 4 H+(aq) + 2 e-

Moles of MnO2 = mass/molar mass

= 1150/86.9368 = 13.228 mol

Moles of electrons required = 2 x moles of MnO2

= 2 x 13.228 = 26.456 mol

Total charge = moles of electrons x Faraday constant

= 26.456 x 96485 = 2.5526 x 106 C

Time = total charge/current

= 2.5526 x 106/30.5 = 8.3692 x 104 s = 23.2 hours