Question

A chemist designs a galvanic cell that uses these two half-reactions:

half-reaction	standard reduction potential
O2 (g) + 4H+ (aq) +4e− →   2H2O (l)	=E0red+1.23V
Zn+2 (aq) +2e− →   Zn (s)	=E0red−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? Yes No If you said it was possible to calculate the cell voltage, do so and enter your answer here. Round your answer to 3 significant digits. V

Answer 1

Balanced equation for the reduction half-reaction that happens at the cathode:

O₂(g) + 4 H⁺(aq) + 4e− 2 H₂O (l)    E⁰_red = +1.23 V ---------- equation (i)

Balanced equation for the reduction half-reaction that happens at the anode:

2 Zn⁺²(aq) + 4 e⁻ 2 Zn (s)      E⁰_red = − 0.763 V

To write it as oxidation half reaction:

2 Zn (s) 2 Zn⁺²(aq) + 4 e⁻                    E⁰_ox = + 0.763 V ----------equation (ii)

Balanced equation for the overall spontaneous reaction (adding equation (i) and equation (ii)) that powers the cell:

2 Zn (s) + O₂(g) + 4 H⁺(aq) 2 Zn⁺²(aq) + 2 H₂O (l)

In spontaneous reaction, the zinc is oxidized as it has positive oxidation potential and oxygen is reduced as it has positive reduction potential.

Yes. We have enough information to calculate the cell voltage under standard conditions .

Standard cell potential or voltage,

E⁰_cell = E⁰_red + E⁰_ox

         = (+1.23) + (+0.763)

         = + 1.99 V