Final answer:
The voltage of the CuCl2 and Zn redox reaction is determined by subtracting the oxidation potential of Zn (effective +0.76 V) from the reduction potential of Cu (0.34 V), resulting in a cell potential of -0.42 V.
Step-by-step explanation:
The potential (voltage) of a redox reaction involving CuCl2 and Zn can be determined using reduction potentials. The standard reduction potential for the half-reaction Zn2+(aq) + 2e- → Zn(s) is -0.76 V. However, for the copper half-reaction Cu2+(aq) + 2e- → Cu(s), the standard reduction potential is +0.34 V. In this redox reaction, zinc is oxidized to Zn2+ and copper is reduced to Cu.
The cell potential (Ecell) of a voltaic cell is calculated by subtracting the anode's potential from the cathode's potential. The oxidation of zinc to Zn2+ is the anode reaction and it has an effective potential of +0.76 V after reversing the sign of its reduction potential. The reduction of Cu2+ to Cu is the cathode reaction, with a potential of +0.34 V. Therefore:
Ecell = Ecathode - Eanode = +0.34 V - (+0.76 V) = -0.42 V
Hence, the cell potential for the reaction CuCl2 + Zn → ZnCl2 + Cu is -0.42 V, indicating that the reaction is non-spontaneous under standard conditions.