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Q. Calculate $E$.$M$.$F$. of following cell at $298 \,K$ $Zn(s) | ZnSO_4(0.01 \,M)||CuSO_4( 1.0 \,M ) |Cu (s)$ if $E^{\circ}$ cell $= 2.0 \,V$

MHT CETMHT CET 2019

Solution:

Key Idea General form of Nernst equation is

$E_{\left(M^{n+} / M\right)}=E_{\left(M^{n+} / M\right)}^{0}-\frac{R T}{n F} \ln \frac{[M]}{\left[M^{n+}\right]}$

Oxidation $Zn ( s ) \longrightarrow Zn ^{2+}(a q)[0.01 M ]+2 e^{-}$

Reduction $Cu ^{2+}(a q)[1.0] M \longrightarrow Cu (s)$

Using Nernst equation,

$E_{\text {cell }} =E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \left[\frac{1}{100}\right] $

$=2-\frac{0.59}{2} \times-2=2+0.059$

$=2.059 V$