A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s) (aq) (aq) (l) (aq) (aq) 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

Reduction (cathode) : Cu²⁺ (⁺aq) + 2 e⁻ → Cu (s)

Oxidation (anode) : Zn (s) → Zn²⁺ (⁺aq) + 2 e⁻

Cu²⁺ (⁺aq) + Zn (s) → Cu (s) + Zn²⁺ (⁺aq)

E°cell = 1.10 V

Explanation:

The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.

Suppose we have the following half-reactions.

Cu²⁺ (⁺aq) + 2 e⁻ → Cu (s) E°red = 0.34 V

Zn²⁺ (⁺aq) + 2 e⁻ → Zn (s) E°red = - 0.76 V

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode) : Cu²⁺ (⁺aq) + 2 e⁻ → Cu (s) E°red = 0.34 V

Oxidation (anode) : Zn (s) → Zn²⁺ (⁺aq) + 2 e⁻ E°red = - 0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺ (⁺aq) + Zn (s) → Cu (s) + Zn²⁺ (⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.