Question

During the discharge of an alkaline battery, 4.30 g of Zn are consumed at the anode of the battery.

a.)What mass of MnO2 is reduced at the cathode during this discharge?

b.) How many coulombs of electrical charge are transferred from Zn to MnO2?

Answer 1

Anode: Zn + 2OH^- -------> ZnO + H2O + 2e-
Cathode: 2MnO2 + H2O + 2e- -------> Mn2O3 + 2OH^-
For the oxidation of 1 mole of Zn, 2 moles of Mn (in the form of MnO2) are reduced.
1 mole of Zn = 65 g/mol
1 mole of MnO2 = 87 g/mol
65 g Zn reduces 2 x 87 g MnO2
4.30 g Zn reduces;
4.30 x 2 x (87 / 65) = 11.51 g MnO2 is reduced at the cathode.

B)

Each zinc atom transfers two electrons
4.30 g * 1 mol Zn / 65.39 g * 2 mol e- / 1 mol Zn = 0.1315 mol e-
1 Mole electrons carry 96485 C
So, coulombs of electrical charge = 0.1315 mol x 96485 C

= 12690 C