Question

Which of the following represent the correct expressions to calculate reduction potential of the following half cells at $25{}^{o}C$.
$C{l}^{-}\left(aq\right)|C{l}_2(g,1atm)|Pt\left(s\right)$

• A. $E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-0.059log\frac{1}{\left[C{l}^{-}\right]}$
• B. $E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0+0.059log\left[C{l}^{-}\right]$
• C. $E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-0.059log\left[C{l}^{-}\right]$
• D. $E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-\frac{0.059}{2}log\left[C{l}^{-}\right]$

Answer 1

The correct option is C $E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-0.059log\left[C{l}^{-}\right]$

In the given cell representation, the gas is indicated before the electrode. Hence, it is a cathode half cell where the reduction reaction occurs.
Thus, the half cell reaction is,

$\frac{1}{2}C{l}_2\left(g\right)+e^{-}\leftrightharpoons C{l}^{-}\left(aq\right)$ (reduction)
(1 atm)

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,

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

$F=96500C$

$T=25+273=298K$, 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 inovolved in the reaction.

$\frac{1}{2}C{l}_2\left(g\right)+e^{-}\leftrightharpoons C{l}^{-}\left(aq\right)$ (reduction)
(1 atm)

$E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-\frac{0.059}{1}log\frac{\left[C{l}^{-}\right]}{[C{l}_2{]}^{\frac{1}{2}}}$


$E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-0.059log\left[C{l}^{-}\right]$