Question

Write the cell reaction and calculate the e.m.f. of the following cell at $298K$ .
$Sn\left( s \right)|S{n^{2 + }}\left( {0.004M} \right)||{H^ + }\left( {0.020M} \right)|{H_2}\left( g \right)\left( {1bar} \right)|Pt\left( s \right).$
(Given: $E{^\circ _{S{n^{2 + }}/Sn}} = - 0.14V$ ).

Answer 1

The reaction which takes place in the cell is called a cell reaction. The right hand reaction electrode is a cathode where reduction occurs and the left hand reaction electrode is anode where oxidation occurs.
Formula used:
Standard e.m.f. formula:
$E{^\circ _{cell}} = E{^\circ _R} - E{^\circ _L}$
Where, $E{^\circ _{cell}}$ is standard electrode potential of the cell, $E{^\circ _L}$ is the standard electrode potential of left hand side of the reaction and $E{^\circ _R}$ is the standard electrode potential of the right hand side of the reaction.
Nernst equation:
$E{^\circ _{cell}} = {E_{cell}} - \dfrac{{RT}}{n}\ln \left[ Q \right]$
Where, ${E_{cell}}$ is the electrode potential at non-standard condition, $E{^\circ _{cell}}$ is electrode potential at standard condition, $T$ is temperature, $R$ is Gas constant, $n$ is number of electrons and $Q$ is reaction quotient.

Complete step by step answer:
We start with writing the half-cell reactions:
Oxidation at anode:
$Sn\left( s \right) \to S{n^{2 + }}\left( {aq} \right) + 2{e^ - }$
Reduction at cathode:
$2{H^ + }\left( {aq} \right) + 2{e^ - } \to {H_2}\left( g \right)$
Now, moving towards the complete cell reactions:
So, the complete cell reaction is as follows:
$Sn\left( s \right) + 2{H^ + }\left( {aq} \right) \to S{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right)$
Given data in the question,
Standard e.m.f. given, $E{^\circ _{S{n^{2 + }}/Sn}} = - 0.14V$
Concentration of $Sn = 0.004M$
Concentration of ${H^ + } = 0.020M$
Pressure, $P = 1bar = 0.987atm$
For calculating the e.m.f. of the cell, we have to find out the standard e.m.f. first,
For finding the standard e.m.f.,
$E{^\circ _{cell}} = E{^\circ _{{H^ + }|{H_2}}} - E{^\circ _{Sn|S{n^{2 + }}}}$
Putting the values of standard electrode potential of the both the sides of the reactions, in order to find out the standard electrode potential of the reaction.
$\Rightarrow E{^\circ _{cell}} = 0.0V - ( - 0.14V) = 0.14V$
So, the standard electrode potential of the cell is $0.14V$ .
Now, for finding the e.m.f. we have to use the Nernst equation,
$E{^\circ _{cell}} = {E_{cell}} - \dfrac{{RT}}{n}\ln \left[ Q \right]$
Now, putting the values of the standard electrode potential, the number of the electrons exchanged is $n$ and different data given,
$\Rightarrow {E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.0592}}{n}\log \dfrac{{\left[ {S{n^{2 + }}} \right] \times {P_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}}$
After putting the values and solving to find the answer.
$\Rightarrow {E_{cell}} = 0.14V - \dfrac{{0.0592}}{2}\log \dfrac{{0.004M \times 0.987atm}}{{{{\left[ {0.020} \right]}^2}}} = 0.111V$
So, the e.m.f. of the given cell at $298K$ is $0.111V$ .

Note:
Nernst equation is an equation which is used to find out the cell potentials of any reactions at non-standard conditions when the reactants concentration, products concentrations, temperature etc are given.