Question

What is the potential of a cell made up of Zn/Zn2+ and Cu/Cu2+ half-cells at 25ºC if [Zn2+] = 0.25 M and [Cu2+] = 0.15 M?

Answer 1

Answer:

Cell potential = +1.09V

Explanation:

Given:

[Zn2+] = 0.25 M

[Cu2+]  = 0.15 M

In this electrochemical cell, Zn/Zn2+ half cell will serve as the anode and Cu/Cu2+ will act as the cathode.

The two half reactions and the reduction potentials based on the standard reduction potential chart are:

Anode (Oxidation):

$Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}$.......E°(anode) = -0.76 V

Cathode(Reduction):

$Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)$........E°(cathode)= +0.34 V

Overall reaction:

$Zn(s) + Cu^{2+}(aq)\rightarrow Cu(s) + Zn^{2+}(aq)$

$E_{cell}^{0} =E_{cathode}^{0}-E_{anode}^{0}=0.34-(-0.76)=1.10V$

Based on the Nernst equation:

$Ecell = E_{cell}^{0}-\frac{0.0592V}{n}log\frac{[Zn2+]}{[Cu2+]}$

$Ecell = 1.10-\frac{0.0592V}{2}log\frac{[0.25]}{[0.15]}=1.09V$