Question
Mark the correct relationship from the following :
A.
Equilibrium constant is related to $$emf$$ as $$\log \,K = \frac{{nFE}}{{2.303\,RT}}$$
B.
$$EMF$$ of a cell $$Zn\left| {Zn_{\left( {{a_1}} \right)}^{2 + }} \right|\left| {Cu_{\left( {{a_2}} \right)}^{2 + }} \right|Cu$$ is $$E = {E^ \circ } - \frac{{0.591}}{n}\log \frac{{\left[ {{a_2}} \right]}}{{\left[ {{a_1}} \right]}}$$
C.
Nernst equation is $${E_{{\text{cell}}}} = $$ $$E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{n}\log \frac{{\left[ {{\text{Products}}} \right]}}{{\left[ {{\text{Reactants}}} \right]}}$$
D.
For the electrode $$\frac{{{M^{n + }}}}{M}$$ at $$298\,K;$$ $$E = {E^ \circ } + \frac{{0.591}}{n}\log \left[ {{M^{n + }}} \right]$$
Answer :
Nernst equation is $${E_{{\text{cell}}}} = $$ $$E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{n}\log \frac{{\left[ {{\text{Products}}} \right]}}{{\left[ {{\text{Reactants}}} \right]}}$$
Solution :
$$\left( {\text{A}} \right)\log \,K = \frac{{nFE_{{\text{cell}}}^ \circ }}{{2.303\,RT}}$$
$$\left( {\text{B}} \right){E_{{\text{cell}}}} = E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{2}\log \frac{{\left[ {{a_1}} \right]}}{{\left[ {{a_2}} \right]}}$$
$$\left( {\text{D}} \right)$$ Expression at $$298\,K,$$ is $$E = {E^ \circ } + \frac{{0.0591}}{n}\log \left[ {{M^{n + }}} \right]$$