Answer to Question #285738 in Physical Chemistry for francisca

Question

The following cell has a potential of +1.2272 V at 25°C: Zn(s)|ZnCl2(aq 0.0005m)||Hg2Cl2(s)|Hg(l)|Pt(s). Determine E for this cell assuming the Debye Hückel equation is satisfied.

Answer 1

Zn_(s)$+$ Hg²⁺_(aq) $\rightleftharpoons\>$ $Zn^{2+}_{(aq)}+Hg_{(l)}$

$E=E^0\>'+\frac{RT}{nF}$ In $\frac{C^*_0}{C^*_R}$

Where $E^0\>'$ (V) is formal potential

$C^*[Mol\>L^{-1}]$ is the bulk concentration for the considered species

$\frac{RT}{nF}=\frac{8.314×298.15}{2×96485}=0.01285$

The activity for a very dilute solution is the same as the concentration

$\therefore\>E=1.2272+0.01285\> In\> \>\frac{0.0005}{1}$

$=1.1295V$