Question

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

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

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 $F=96485Cmo{l}^{-1}$
We get,
$E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.0591}{n}\log Q$
where, $Q=\frac{{\text{ ion }}_{\text{RHS }}}{{\text{ ion }}_{\text{LHS }}}$
$E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.0591}{n}\log \frac{io{n}_{RHS}}{{\text{ ion }}_{LHS}}$