Question

Consider the following equation for a cell reaction
$A+B \rightleftharpoons C+D ; \,\,\,\,\, E^{\circ}=x$ volt $, K_{e q}=K_{1}$
$2 A+2 B \rightleftharpoons 2 C+2 D ; \,\,\, E^{\circ}=y$ volt $, K_{e q}=K_{2}$ then

• A. $x=y, K_{1}=\frac{1}{K_{2}}$
• B. $ x=y, K_{1}=K_{2}^{2}$
• C. $ x=y, K_{2}=K_{1}^{2}$
• D. $x=2 y, K_{1}=2 K_{2}$

Answer 1

Answer: (C)

$K_{1}=\frac{[C][D]}{[A][B]} ; K_{2}=\frac{[C]^{2}[D]^{2}}{[A]^{2}[B]^{2}},$ therefore $K_{2}=K_{1}^{2}$

Since, on multiplying equations by any stoichiometric coefficient, standard electrode potential value does not get multiplied, therefore $x=y$