Question

Calculate using Nernst equation the cell potential of the following electrochemical cell at ${\text{298K}}$.
${\text{Zn}}\left( {\text{s}} \right)\left| {{\text{Z}}{{\text{n}}^{2 + }}} \right.\left( {{\text{aq}}} \right)\left\| {{\text{S}}{{\text{n}}^{2 + }}} \right.\left( {{\text{aq}}} \right)\left| {{\text{Sn}}} \right.\left( {\text{s}} \right)$
$\left( {0.04{\text{M}}} \right)$ $\left( {0.03{\text{M}}} \right)$ .Given ${{\text{E}}^ \circ } = - 0.76{\text{V}},{{\text{E}}^ \circ } = - 0.14{\text{V}}$

Answer 1

An electrochemical system consists of a vessel containing an electrolyte into which two electrodes are dipped. In an electrolytic cell, a flow of current produces a chemical reaction. This process involves transfer of electrons which changes the oxidation state of the molecule or ions. Electrochemical cells are of two types-electrolytic and voltaic cells.
Nernst equation is the principle behind this problem.

Formula used:

Complete answer: or Complete step by step answer:
The electrodes in the electrochemical cells get oxidized or reduced. Electrochemical cells consists of two half cells. Each half cell has an electrode and an electrolyte. Species from one half (anode) lose electrons (oxidation) and species from other half (cathode) gain electrons (reduction).
In the cell diagram, first part denotes the anodic part while the last part is the cathodic part.
In the cell diagram given, ${\text{Zn}}\left( {\text{s}} \right)\left| {{\text{Z}}{{\text{n}}^{2 + }}} \right.\left( {{\text{aq}}} \right)\left\| {{\text{S}}{{\text{n}}^{2 + }}} \right.\left( {{\text{aq}}} \right)\left| {{\text{Sn}}} \right.\left( {\text{s}} \right)$, zinc get oxidized to ${\text{Z}}{{\text{n}}^{2 + }}$ and ${\text{S}}{{\text{n}}^{2 + }}$ get reduced to ${\text{Sn}}$.
It is given that the concentration of ${\text{Z}}{{\text{n}}^{2 + }}$, $\left[ {{\text{Z}}{{\text{n}}^{2 + }}} \right] = 0.04{\text{M}}$ and that of ${\text{S}}{{\text{n}}^{2 + }}$, $\left[ {{\text{S}}{{\text{n}}^{2 + }}} \right] = 0.03{\text{M}}$
Cell potential at anode, ${{\text{E}}^ \circ }_{{\text{Zn}}\left| {{\text{Z}}{{\text{n}}^{2 + }}} \right.} = - 0.76{\text{V}}$ and cell potential at cathode, ${{\text{E}}^ \circ }_{{\text{Sn}}\left| {{\text{S}}{{\text{n}}^{2 + }}} \right.} = - 0.14{\text{V}}$
Standard potential, ${{\text{E}}^ \circ } = {{\text{E}}^ \circ }_{{\text{cathode}}} - {{\text{E}}^ \circ }_{{\text{anode}}} \Leftrightarrow {{\text{E}}^ \circ } = {{\text{E}}^ \circ }_{{\text{S}}{{\text{n}}^{2 + }}\left| {{\text{Sn}}} \right.} - {{\text{E}}^ \circ }_{{\text{Zn}}\left| {{\text{Z}}{{\text{n}}^{2 + }}} \right.}$
Substituting the values, we get
${{\text{E}}^ \circ } = - 0.14{\text{V}} - \left( { - 0.76{\text{V}}} \right) = + 0.62{\text{V}}$
Now we can use the Nernst equation for finding the cell potential.
${\text{E}} = {{\text{E}}^ \circ } - \dfrac{{0.0592}}{{\text{F}}}\log \dfrac{{\left[ {{\text{Z}}{{\text{n}}^{2 + }}} \right]}}{{\left[ {{\text{S}}{{\text{n}}^{2 + }}} \right]}}$
Substituting the values, we get
${\text{E}} = 0.62{\text{V}} - \dfrac{{0.0592}}{2}\log \dfrac{{\left[ {0.04} \right]}}{{\left[ {0.03} \right]}}$
On simplification, we get
${\text{E}} = 0.62{\text{V}} - 0.0296\log 1.2$
${\text{E}} = 0.62{\text{V}} - \left( {3.7 \times {{10}^{ - 3}}} \right)$
Solving,
${\text{E}} = 0.62{\text{V}} - \left( {3.7 \times {{10}^{ - 3}}} \right) = 0.616{\text{V}}$

Note:
Gibbs’ free energy for a cell can also be calculated using Nernst equation. The Nernst equation allows us to calculate potential when the two cells are not in $1{\text{M}}$ concentration. At equilibrium, forward and reverse reactions occur at equal rates. Thus the cell potential is zero volts. Equilibrium constant can be calculated from the cell potential.