Question

Calculate the electrode potential at a copper electrode dipped in a 0.1M solution of copper sulphate at ${25^ \circ }$C. The standard electrode potential of ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/Cu}}$ system is 0.34 volt at 298K.

Answer 1

The reduction electrode potential or electrode potential of an electrode is given by Nernst equation. Nernst equation for a single electrode potential is ${E_{\left( {cell} \right)}} = {E^ \circ } - \dfrac{{RT}}{{nF}}\ln Q$ .

Complete step by step answer:
To calculate the electrode potential of any given electrolytic cell, we use the Nernst equation. The Nernst equation is used to calculate even the electrode potential of a single electrode. The formula to calculate single electrode potential is given as follows.
${E_{\left( {red} \right)}} = {E^ \circ }_{\left( {{M^{n + }}/M} \right)} - \left[ {\dfrac{{2.303RT}}{{nF}}} \right]\log \left[ {\dfrac{1}{{\left[ {{M^{n + }}} \right]}}} \right]$ where,
${E_{\left( {red} \right)}}$ is the reduction potential of a single electrode potential which is to be calculated for copper electrodes.

${E^ \circ }_{\left( {{M^{n + }}/M} \right)}$ is the standard reduction potential of the given metal ion in the solution, which is 0.34 for copper metal to copper ion.
$\left[ {{M^{n + }}} \right]$ is the concentration of the metal ion solution, given as 0.1M.
R is the universal gas constant, to be taken as 8.314 J/K mol in SI units.
T is the temperature in kelvin, which is given as 298K.
n is the number of moles of electrons flowing from one mole of the metal into the solution, which is 2 for copper in $C{u^{ + 2}}/Cu$ system
F is the faraday constant and its value for one mole of electrons is 96500 Coulomb /mol.

Let us now substitute the values of all the terms in the above given Nernst equation, of single electrode potential. As the values of ‘R’, temperature (T), faraday constant (F) and 2.303 are constant, the term $\left[ {\dfrac{{2.303RT}}{{nF}}} \right]$ is calculated to be $\left[ {\dfrac{{0.0591}}{n}} \right]$ in the equation. Nernst equation is also written as ${E_{\left( {Cell} \right)}} = {E^ \circ }_{\left( {C{u^{ + 2}}/Cu} \right)} - \dfrac{{\left[ {0.0591} \right]}}{n}\log \left[ {\dfrac{1}{{\left[ {C{u^{ + 2}}} \right]}}} \right] {E_{\left( {cell} \right)}} = 0.34 - \left[ {\dfrac{{0.0591}}{2}} \right]\log \left[ {\dfrac{1}{{\left[ {0.1} \right]}}} \right] {E_{\left( {cell} \right)}} = 0.34 - 0.02955 \times 1\left( {\because {{\log }_{10}}\left( {10} \right) = 1} \right) {E_{\left( {cell} \right)}} = 0.3104V$
So the electrode potential of a copper electrode dipped in a 0.1M solution of copper sulphate at ${25^ \circ }$C is 0.03104 Volts.

Note: The standard reduction potentials of any metal is the reduction potential obtained at standard conditions of pressure and temperature. They can be known from referring to the reduction potential chart. Most of the time its values are given in the numerical based questions.