Question

What will be the value of $\log K_{eq}$ for the given reaction at $298K$?
where,
$K_{eq}\text{is the equilibrium constant for the given reaction}$

$Cu\left(s\right)+2A{g}^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2Ag\left(s\right);E_{cell}^0=0.46V$

• A. $15.6$
• B. $8.2$
• C. $4.9$
• D. $23.9$

Answer 1

The correct option is A $15.6$

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

Nernst equation is,
$E=E^0-\frac{2.303RT}{nF}\log \left(\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}\right).....(Eqn.1)$

Substituting,

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

$F=96500C/mol$

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

$E_{cell}=E_{cell}^0-\frac{0.059}{n}\log \left(\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}\right)....(Eqn.2)$
where,
n is the number of electrons inovolved in the reaction.

$Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}$
Q is reaction quotient

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

Hence, nernst equation becomes
$0=E_{cell}^0-\frac{0.059}{n}log{K}_{eq}$

$\log K_{eq}=\frac{n}{0.059}{E}_{cell}^0$

For given cell reaction,
$Cu\left(s\right)+2A{g}^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2Ag\left(s\right)$
$E^0=0.46V$
$\therefore$
$\log K_{eq}=\frac{n}{0.059}\times 0.46$

$\log K_{eq}=\frac{2\times 0.46}{0.059}$

$\log K_{eq}\approx 15.6$