Question

A chemist designs a galvanic cell that uses these two half-reactions: MnO_(4)^(-)+8H^(+)(aq)+5e^(-)->Mn^(2+)(aq)+4H_(2)O(l) Eo=+1.51 V. Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answer 1

The half-reaction that happens at the cathode is the reduction of MnO4− to Mn2+:

MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)

The half-reaction that happens at the anode is the oxidation of Zn to Zn2+:

Zn(s) → Zn2+(aq) + 2e−

The balanced equation for the overall reaction that powers the cell is obtained by multiplying the anode half-reaction by 5 and adding it to the cathode half-reaction:

5Zn(s) + MnO4−(aq) + 8H+(aq) → 5Zn2+(aq) + Mn2+(aq) + 4H2O(l)

This reaction is spontaneous as written because the cell voltage under standard conditions is positive. To calculate the cell voltage, we can use the formula:

Ecell = Ered,cathode − Ered,anode

Substituting the standard reduction potentials given in the question, we get:

Ecell = 1.51 V − (−0.763 V)

Ecell = 2.273 V

Therefore, the cell voltage under standard conditions is 2.273 V. I hope this answer helps you understand how to write balanced equations for galvanic cells and calculate their cell voltages. Thank you for chatting with me.