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
