Question

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

Answer 1

Approximately
`1.10\; {\rm V}` under standard conditions.

Step-by-step explanation:

Equation for the overall reaction:

`{\rm CuCl_(2)}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_(2)} \, (aq) + {\rm Cu}\, (s)`.

Write down the ionic equation for this reaction:

`\begin{aligned}& {\rm Cu^(2+)}\, (aq) + 2\; {\rm Cl^(-)}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^(2+)} \, (aq) + 2\; {\rm Cl^(-)}\, (aq) + {\rm Cu}\, (s)\end{aligned}`.

The net ionic equation for this reaction would be:

`{\rm Cu^(2+)}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^(2+)}\, (aq) + {\rm Cu}\, (s)`.

In this reaction:

• Zinc loses electrons and was oxidized (at the anode):
  `{\rm Zn}\, (s) \to {\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)}`.
• Copper gains electrons and was reduced (at the cathode):
  `{\rm Cu^(2+)}\, (aq) + 2\, {\rm e^(-)} \to {\rm Cu} \, (s)`.

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that
`{\rm Zn}\, (s) \to {\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)}` is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction,
`{\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)} \to {\rm Zn}\, (s)`, is reduction and is likely on the table.

• 
  `E(\text{anode}) = -0.7618\; {\rm V}` for
  `{\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)} \to {\rm Zn}\, (s)`, and
• 
  `E(\text{cathode}) = 0.3419\; {\rm V}` for
  `{\rm Cu^(2+)}\, (aq) + 2\, {\rm e^(-)} \to {\rm Cu} \, (s)`.

The reduction potential of
`{\rm Zn}\, (s) \to {\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)}` would be
`-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}`, the opposite of the reverse reaction
`{\rm Zn^(2+)}\, (aq) + 2\, {\rm e^(-)} \to {\rm Zn}\, (s)`.

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

`\begin{aligned} E^(\circ) &= E^(\circ)(\text{cathode}) + (-E^(\circ)(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}`.