A galvanic cell is powered by the following redox reaction:

3Br2 (l) + 2MnO2 (s) + 8OH^ (-) (aq) = 6Br^ (-) (aq) + 2MnO4^ (-) (aq) + 4H2O (l)

a. Write a balanced equation for the half-reaction that takes place at the cathode.

b. Write a balanced equation for the half-reaction that takes place at the anode.

c. Calculate the cell voltage under standard conditions.

3Br2 (l) + 2MnO2 (s) + 8OH^ (-) (aq) - -> 6Br^ (-) (aq) + 2MnO4^ (-) (aq) + 4H2O (l)

The cathode is where reduction takes place and oxidation takes place at the anode.

a. Write a balanced equation for the half-reaction that takes place at the cathode.

This means we have to write out the reduction reaction here.

The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction.

3Br2 (l) - -> 6Br^ (-) (aq)

Reduction is the gain of electrons so we have;

3Br2 (l) + 6e^ (-) - -> 6Br^ (-) (aq)

Br2 (l) + 2e^ (-) - -> 2Br^ (-) (aq)

b. Write a balanced equation for the half-reaction that takes place at the anode.

2MnO2 (s) + 8OH^ (-) (aq) - -> 2MnO4^ (-) (aq) + 4H2O (l)

MnO2 (s) + 4OH^ (-) (aq) - -> MnO4^ (-) (aq) + 2H2O (l) + 3e^ (-)

c. Calculate the cell voltage under standard conditions.

We do this by Looking up the standard potential for the half-reactions.

Reduction - -> + 1.07

Oxidation - -> - 1.70

Add the cell potentials to get the overall standard cell potential.

E⁰cell = E⁰ox + E⁰red = - 1.70 + 1.07 = - 0.63 V