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

[tex]\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}[/tex].

The net ionic equation for this reaction would be:

[tex]{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)[/tex].

In this reaction:

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

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

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

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

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

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

[tex]\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}[/tex].