Question

The standard free energy change for the given reaction is - $10.26kcalmo{l}^{-1}$ at ${25}^{\circ }C.$
$H_2\left(g\right)+2AgCl\left(s\right)\to 2Ag\left(s\right)+2H^{+}\left(aq\right)+2C{l}^{-}\left(aq\right)$
A cell using the above reaction is operated at ${25}^{\circ }C$ under conditions,
$P_{H_2}=1atm$
$\left[H^{+}\right]\text{and}\left[C{l}^{-}\right]=0.1M$

Calculate the emf of the cell.

• A. $0.140V$
• B. $0.240V$
• C. $0.340V$
• D. $0.440V$

Answer 1

The correct option is C $0.340V$

For the given reaction,
$H_2\left(g\right)+2AgCl\left(s\right)\to 2Ag\left(s\right)+2H^{+}\left(aq\right)+2C{l}^{-}\left(aq\right)$
Number of electrons involved in the reaction is 2
$\Delta G^0=-10.26kcalmo{l}^{-1}$
Since,
$\Delta G^0=-nF{E}^0$
$E^0=-\frac{10.26\times {10}^3\times 4.184}{-2\times 96500}\because 1cal=4.184J$
$E^0=0.2224V$
$E^0\approx 0.22V$

Nernst equation for
$aA+bB\leftrightharpoons cC+dD$

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

Substituting,

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

$F=96500C/mol$

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

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

Hence, for given reaction
$H_2\left(g\right)+2AgCl\left(s\right)\to 2Ag\left(s\right)+2H^{+}\left(aq\right)+2C{l}^{-}\left(aq\right)$

$E_{cell}=E_{cell}^0-\frac{0.059}{n}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}}$

$E_{cell}=E_{cell}^0-\frac{0.059}{2}log\frac{\left[0.1{]}^2\right[0.1{]}^2}{1}$

$E_{cell}=E_{cell}^0-\frac{0.059}{2}log{10}^{-4}$
$E_{cell}=E_{cell}^0+\frac{0.059\times 4}{2}$
$E_{cell}=0.22+0.118$
$E_{cell}=0.338$
$E_{cell}\simeq 0.34V$