Question

Calculate the potential of Ag-Zn cell at 298 K if the concentration of Ag+ and Zn2+ are 5.2 x 10-6 M and 1.3 x 10-3 M respectively. Eo of the cell at 298 K is 1.56 V. Write the cell representation and reactions. Calculate the change in free energy ∆G for the reduction of 1 mole of Ag+.

Answer 1

Cell representation of this cell:

$Zn | Zn^{+2} || Ag^{+1} | Ag$

Cell reactions:

$Zn \to Zn^{+2}+2e^-$

$Ag^++e^- \to Ag$

Net reaction:$Zn +2Ag^+ \to Zn^{+2} +2Ag$

Number of electrons used$(n)=2$

$E_{cell}=E_0-\frac{0.0591}{n} \space log_{10} \frac{[Zn^{+2}]}{[Ag^+]^2}$

$E_{cell}=1.56-\frac{0.0591}{2} \space log_{10} \frac{1.3 \times 10^{-3}}{(5.2 \times 10^{-6})^2}$

$E_{cell}=1.56-0.227=1.33 V$

Gibbs energy change$=nFE=2 \times 96500 \times 1.33=256,690 J=256.7 KJ$