Question

A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Zn2+ solutions, 0.129 M and 0.427 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Zn2+ has changed by 0.047 M at 291 K?
Zn2+ + 2 e- ? Zn Eo = -0.761 V

Answer 1

Explanation:

The given reaction at cathode will be as follows.

At cathode: $Zn^{2+} + 2e^{-} \rightarrow Zn$,     $E_{o}$ = -0.761 V

At anode: $Zn \rightarrow Zn^{2+} + 2e^{-}$,       $E_{o}$ = 0.761

Therefore, net reaction equation will be as follows.

                 $Zn^{2+} + Zn \rightarrow Zn + Zn^{2+}$

Initial:     0.129         -            -       0.427

Change:  -0.047      -            -     -0.047

Equilibrium: (0.129 - 0.047)      (0.427 - 0.047)

                = 0.082                       = 0.38

As $E^{o}_{cell}$ for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

                   E_{cell} = $E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}$

                              = $0 - \frac{8.314 atm L/mol K \times 291 K}{2 \times 96500} ln \frac{0.38}{0.082}$

                              = 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.