What is the voltage of the cell represented at Zn(s) Zn 2+(aq) /Cu2+(aq) /Cu(s) given that for Cu 2+(aq)/ Cu (s) Eo = 0.337V and for Zn2+(aq) /Zn(s)' E o = ...

What is the voltage of the cell represented at Zn_(s) Zn ²⁺_(aq) /Cu²⁺_(aq) /Cu_(s) given that for Cu ²⁺_(aq)/ Cu _(s) E_o = 0.337V and for Zn²⁺_(aq) /Zn_(s)' E _o = -0.763V?

Answer Details

The voltage of the cell can be calculated using the formula: E_cell = E_{reduction, cathode} - E_{oxidation, anode} The half-reactions for the cell are: Cathode: Cu²⁺ (aq) + 2e^- → Cu(s) (E_{reduction, cathode} = 0.337V) Anode: Zn(s) → Zn²⁺ (aq) + 2e^- (E_{oxidation, anode} = -0.763V) Substituting the values into the formula, we get: E_cell = 0.337V - (-0.763V) = 1.1V Therefore, the voltage of the cell represented at Zn_(s) Zn ²⁺_(aq) /Cu²⁺_(aq) /Cu_(s) is +1.10V. Answer: +1.10V