Question

Calculate the reduction potential for the following half cells at ${25}^{o}C$.
(i) $Mg\left(s\right)|M{g}^{2+}(aq,1\times {10}^{-4}M);E_{Mg/M{g}^{{}^{2+}}}^0=+2.36V$
(ii) $C{l}_2\left(g\right)|C{l}^{-}(aq,2\times {10}^{-5}M);E_{C{l}_2/C{l}^{-}}^0=+1.36V$

[$log2=0.301$]

• A. $E_{M{g}^{2+}/Mg}=1.34V$
  $E_{C{l}_2/C{l}^{-}}=-1.16V$
• B. $E_{M{g}^{2+}/Mg}=-1.34V$
  $E_{C{l}_2/C{l}^{-}}=+1.16V$
• C. $E_{M{g}^{2+}/Mg}=2.48V$
  $E_{C{l}_2/C{l}^{-}}=1.16V$
• D. $E_{M{g}^{2+}/Mg}=-2.48V$
  $E_{C{l}_2/C{l}^{-}}=1.64V$
  

Answer 1

The correct option is D $E_{M{g}^{2+}/Mg}=-2.48V$

$E_{C{l}_2/C{l}^{-}}=1.64V$

(i)
Given,
The standard oxidation potential of the electrode $Mg/M{g}^{2+}$ is $2.36V$.
The reduction potential of the electrode $M{g}^{2+}/Mg$ is $-2.36V$.
The reduction reaction of the electrode is
$M{g}^{2+}\left(aq\right)+2e\rightleftharpoons Mg\left(s\right)\left(reduction\right)$

Nernst equation for the half cell is

$E_{M^{n+}/M}=E_{M^{n+}/M}^0-\frac{2.303RT}{nF}log\frac{\left[M\right(s\left)\right]}{\left[M^{n+}\right(aq\left)\right]}$

n is the number of electrons involved in the half cell reaction.
$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^o-\frac{0.0591}{2}log\frac{1}{{10}^{-4}}$

$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^o+\frac{0.0591}{2}log{10}^{-4}$

$E_{M{g}^{2+}/Mg}=-2.36-\frac{4\times 0.0591}{2}$
$E_{M{g}^{2+}/Mg}=-2.48V$


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

Similarly for the above reaction,
$E_{C{l}_2/C{l}^{-}}=E_{C{l}_2/C{l}^{-}}^0-\frac{0.059}{1}log\frac{\left[C{l}^{-}\right]}{[C{l}_2{]}^{\frac{1}{2}}}$

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

$E_{C{l}_2/C{l}^{-}}=1.36-\frac{0.0591}{1}log(2\times {10}^{-5})$

$E_{C{l}_2/C{l}^{-}}=1.36-\frac{0.0591}{1}\times 0.301+\frac{5\times 0.0591}{1}$'
$E_{C{l}_2/C{l}^{-}}=1.64V$