Question

At $25\,{}^{\circ}C$, Nernst equation is

• A. $E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}log \frac{\left[ion\right]_{RHS}}{\left[ion\right]_{LHS}}$
• B. $E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}log \frac{\left[M\right]_{RHS}}{\left[M\right]_{LHS}}$
• C. $E_{cell}=E^{\circ}_{cell}+\frac{0.0591}{n}log \frac{\left[ion\right]_{RHS}}{\left[ion\right]_{LHS}}$
• D. $E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}log \frac{\left[ion\right]_{LHS}}{\left[ion\right]_{RHS}}$

Answer 1

Answer: (A)

Nernst eauation is given by,
$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{ RT }{ nF } \ln Q$
At $T =25^{\circ} C$, on substituting values of $R$ is $8.314 / mol$.
$K$ and $F =96485\, C\, mol ^{-1}$
We get,
$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.0591}{ n } \log Q$
where, $Q =\frac{\text { ion }_{\text {RHS }}}{\text { ion }_{\text {LHS }}}$
$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.0591}{ n } \log \frac{ ion _{ RHS }}{\text { ion }_{ LHS }}$