Question

Consider the following equation for a cell reaction
${\text{A}}\, + \,{\text{B}}\,\, \rightleftharpoons {\text{C}}\, + \,{\text{D}}$ ${{\text{E}}^0}\, = \,{\text{X volt}}$, ${{\text{k}}_{{\text{eq}}}}\,{\text{ = }}{{\text{K}}_{\text{1}}}$
${\text{2A}}\, + \,2{\text{B}}\,\, \rightleftharpoons 2{\text{C}}\, + \,2{\text{D}}$ ${{\text{E}}^0}\, = \,{\text{Y volt}}$, ${{\text{k}}_{{\text{eq}}}}\,{\text{ = }}{{\text{K}}_2}$ then:
A) ${\text{x = y}}$, ${{\text{k}}_1}\,{\text{ = }}{{\text{K}}_2}$
B) ${\text{x = 2}}\,{\text{y}}$, ${{\text{k}}_1}\,{\text{ = 2}}\,{{\text{K}}_2}$
C) ${\text{x = y}}$, ${\text{k}}_1^2\,{\text{ = }}{{\text{K}}_2}$
D) ${{\text{x}}^2}{\text{ = y}}$, ${\text{k}}_1^2\,{\text{ = }}{{\text{K}}_2}$

Answer 1

We will write the equilibrium constant expression for each reaction. Then put both the equilibrium constant equal to determine the relation between them. Reduction potential is the potential required for the reduction of a species. Reduction potential does not depend upon stoichiometry.

Complete answer:
The equilibrium constant is defined as the product of the concentration of products having a stoichiometric coefficient as power divided by the product of the concentration of reactants having a stoichiometric coefficient as power.
${\text{A}}\, + \,{\text{B}}\,\, \rightleftharpoons {\text{C}}\, + \,{\text{D}}$ ${{\text{E}}^0}\, = \,{\text{X volt}}$, ${{\text{k}}_{{\text{eq}}}}\,{\text{ = }}{{\text{K}}_{\text{1}}}$
The equilibrium constant express ${{\text{k}}_{{\text{eq}}}}\,{\text{ = }}{{\text{K}}_2}$ ion for the above reaction is as follows:
$\Rightarrow {{\text{k}}_{{\text{eq}}}} = \,\dfrac{{\left[ {\text{C}} \right]\,\,\left[ {\text{D}} \right]}}{{\left[ {\text{A}} \right]\left[ {\text{B}} \right]}}\, = \,{{\text{K}}_{\text{1}}}$…..$(1)$
${\text{2A}}\, + \,2{\text{B}}\,\, \rightleftharpoons 2{\text{C}}\, + \,2{\text{D}}$ ${{\text{E}}^0}\, = \,{\text{Y volt}}$, then:

The equilibrium constant expression for the above reaction is as follows:
$\Rightarrow {{\text{k}}_{{\text{eq}}}} = \,\dfrac{{{{\left[ {\text{C}} \right]}^2}\,\,{{\left[ {\text{D}} \right]}^2}}}{{{{\left[ {\text{A}} \right]}^2}{{\left[ {\text{B}} \right]}^2}}} = \,{{\text{k}}_2}$…..$(2)$
From equation $(1)$ and $(2)$,
$\Rightarrow {\text{k}}_1^2\,{\text{ = }}{{\text{K}}_2}$
The reduction potential of a metal calculated in one molar solution at $298$k temperature is known as standard reduction potential. The stoichiometry does not affect the standard reduction potential because reduction potential is an intensive property. Intensive properties are the properties that do not depend upon mass or numbers of atoms.
So, ${\text{x = y}}$

Therefore, option (C) ${\text{x = y}}$, ${\text{k}}_1^2\,{\text{ = }}{{\text{K}}_2}$ is correct.

Note: If we multiply a reaction with a coefficient then the new equilibrium constant will have that coefficient as power. The equilibrium constant for the backward direction is the inverse of the equilibrium constant for the forward direction. Standard reduction potential is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. For the calculation of reduction potential of cell, half-cell reaction is multiplied with the coefficient to balance the number of electrons exchanged but the standard reduction potential is not multiplied with that coefficient.