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.

Explanation:

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

16

explanation:

the marks go by 4s

c

explanation: