Question

For the galvanic cell,
$Zn ( s )+ Cu ^{2+}(0.02 M ) \rightarrow Zn ^{2+}(0.04 M )+ Cu ( s )$
$E _{\text {cell }}=$ ____$\times 10^{-2}$(Nearest integer)
[Use : $E _{ Cu / Cu ^{2+}}^{0}=-0.34 V , E _{ Zn / Zn ^{2+}}^{0}=+0.76 V,\frac{2.303 RT }{ F }=0.059 \,V$ ]

Answer 1

$Zn _{( s )}+\underset{0.02 M }{ Cu _{( aq .)}^{+2}} \rightarrow \underset{0.04 M }{ Zn ^{+2}}+ Cu ( s )$
Nernst equation $= F _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[2 n ^{+2}\right]}{\left[ Cu ^{+2}\right]}$
$\Rightarrow E _{\text {cell }}\left[ E _{\text {cell }}^{\circ}- E _{ Zn ^{+2} / Zn }^{\circ}\right]-\frac{0.059}{2} \log \frac{0.04}{0.02}$
$\Rightarrow E _{\text {cell }}[0.34-(-0.76)]-\frac{0.059}{2} \log ^{2}$
$\Rightarrow E _{\text {cell }} 1-1-\frac{0.059}{2} \times 0.3010$
$=1.0911=109.11 \times 10^{-2}$
$=109$