Consider the following equation for a cell reaction A+B leftharpoons C+D ; E°=x volt , Ke q=K1 2 A+2 B leftharpoons 2 C+2 D ; E°=y volt , Ke q=K2 then

Question

Consider the following equation for a cell reaction A+B leftharpoons C+D ; E°=x volt , Ke q=K1 2 A+2 B leftharpoons 2 C+2 D ; E°=y volt , Ke q=K2 then (A) x=y, K1=(1/K2) (B)  x=y, K1=K22 (C)  x=y, K2=K12 (D) x=2 y, K1=2 K2

Tardigrade Admin · Accepted Answer

K1=([C][D]/[A][B]) ; K2=([C]2[D]2/[A]2[B]2), therefore K2=K12 Since, on multiplying equations by any stoichiometric coefficient, standard electrode potential value does not get multiplied, therefore x=y

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

$x = y, K_{1} = \frac{1}{K _{2}}$

$x = y, K_{1} = K_{2}^{2}$

$x = y, K_{2} = K_{1}^{2}$

$x = 2 y, K_{1} = 2 K_{2}$

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

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

Q. Consider the following equation for a cell reaction A+B⇌C+D;E∘=x volt ,Keq​=K1​ 2A+2B⇌2C+2D;E∘=y volt ,Keq​=K2​ then

x=y,K1​=K2​1​

x=y,K1​=K22​

x=y,K2​=K12​

x=2y,K1​=2K2​