Question

Which of the following expression(s) is/are correct at equilibrium condition of Daniell cell at ${25}^{\circ }C?$
For Daniel cell, $E_{cell}^0=1.1V$

• A. $\frac{2.303RT}{F}log{K}_{eq}=0.55$
• B. $\frac{2.303RT}{2F}log{K}_{eq}=1.1$
• C. $log{K}_{eq}=\frac{2.2}{0.059}$
• D. $log{K}_{eq}=\frac{0.55}{0.059}$

Answer 1

Answer: (B), (C)

$log{K}_{eq}=\frac{2.2}{0.059}$

For a general electrochemical reaction,
$aA+bB\leftrightharpoons cC+dD$

Nernst equation is,
$E=E^0-\frac{2.303RT}{nF}log\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}.....(Eqn.1)$
$Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}$
Q is reaction quotient
n is number of electron involved in the reaction

At equilibrium,
$Q=K_{eq}$
$\Delta G=0$
$E_{cell}=0$

Thus,
$E_{cell}^0=\frac{2.303RT}{nF}log{K}_{eq}....eqn.2$

For Daniell cell,
$Zn\left(s\right)+C{u}^{2+}\left(aq\right)\to Cu\left(s\right)+Z{n}^{2+}\left(aq\right)$
n = 2
$E_{cell}^0=1.1V$
Substituting in eqn. 2 we get,
$1.1=\frac{2.303RT}{2F}log{K}_{eq}......eqn.3$

Substituting in eqn.3,

$R=8.314J{K}^{-1}mo{l}^{-1}$

$F=96500C/mol$

$T=25+273=298K$, we get,

$1.1=\frac{0.059}{2}log{K}_{eq}$
$log{K}_{eq}=\frac{2.2}{0.059}$

Hence, (b) and (c) are correct expressions.