Question

A galvanic cell consists of a strip of cobalt metal dipping into 1.0 M $C{o}^{2+}$ ions, and another half cell in which a piece of platinum dips into a 1.0 M solution of $C{l}^{-}$ ions. Now $C{l}_2\left(g\right)$ at a pressure of 1.0 atm is passed into this solution. The observed cell voltage is 1.63 V, and as the cell reaction proceeds, the Co electrode becomes negative.
$E^{\ominus }(1/2C{l}_2+e^{-}\to C{l}^{\ominus })=1.36V$. The $E^{\ominus }$ of cobalt electrode is:

• A. $+0.27V$
• B. $+0.54V$
• C. $-0.27V$
• D. none of these

Answer 1

Answer: (C)

$-0.27V$

$E_{cell}=1.63V{E}_{{Cl}_2/{Cl}^{-}}^o=1.36V{E}_{cell}=(1.36-E_{{Co}^{3+}/Co}^o)-\frac{0.0591}{2}log\frac{{\left[{Cl}^{-}\right]}^2}{\left[C{o}^{2+}\right]}{E}_{cell}=1.36-E_{{Co}^{3+}/Co}^o-\frac{0.0591}{1}log\frac{1}{1}1.63=1.36--E_{{Co}^{3+}/Co}^o-0E_{{Co}^{3+}/Co}^o=1.36-1.63=-0.27V$