Question

A cell containing two H electrodes. The negative electrode is in contact with a solution of $ {10^{ - 6}}M$ ${H^ + }$ ion. The e.m.f of the cell is $0.118$ volt at ${25^\circ }C$ . $\left[ {{H^ + }} \right]$ at the positive electrode is :
A. ${10^{ - 4}}$
B. ${10^{ - 5}}$
C. ${10^{ - 3}}$
D. None of the above

Answer 1

Hint: The electromotive force(e.m.f) is defined as the maximum potential difference between two electrodes of galvanic/voltaic cells. The reaction is said to be spontaneous when e.m.f of a cell is positive and the reaction is said to be non-spontaneous when the emf of a cell is negative. For standard hydrogen electrodes, its standard e.m.f is zero.

Complete step by step answer:
Given is that a cell contains two hydrogen electrodes. The cell can be represented as,
$Pt,{H_2}({10^{ - 6}}M)/{H^ + }//{H^ + }/{H_2}(x)Pt$
If the electrode is hydrogen, the platinum foil is used since platinum acts as a surface in which reaction takes place.
Let “x” be the concentration of the positive hydrogen electrode.
The e.m.f of the cell can be calculated by using the Nernst equation.
$E = {E^\circ } - \dfrac{{RT}}{{nF}}\log K......(1)$
Where,
E-emf of the cell
${E^\circ }$ -standard emf
R-gas constant
T-temperature
n-number of electrons involved
F-faraday’s constant
K-reaction quotient
For standard hydrogen electrode, ${E^\circ } = 0$
The electrons involved between two hydrogen electrodes, $n = 2$
Substituting all the constant values such as F,R and thus, the term $\dfrac{{RT}}{{nF}}$ can be written as
$\dfrac{{RT}}{{nF}} = 0.02955$
$(1) \to 0.118V = - \dfrac{{0.0591}}{2} \times \log \left( {\dfrac{{{{(x)}^2}}}{{{{({{10}^{ - 6}})}^2}}}} \right)$
$ \Rightarrow 0.118 = \dfrac{{0.0591}}{1}\log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)$
$ \Rightarrow \dfrac{{0.0591}}{{0.118}} = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)$
$ \Rightarrow \dfrac{{0.118}}{{0.0591}} = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)$
$ \Rightarrow 1.997 = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)$
$ \Rightarrow 99.31 = \dfrac{x}{{{{10}^{ - 6}}}}$
$ \Rightarrow 99.31 \times {10^{ - 6}} = x$
Since $99.31 \approx 100$ , then
$ \Rightarrow {10^{ - 4}} = x$

Thus, the correct answer is option A.

Note: Generally the reaction is said to be spontaneous only if Gibbs free energy, $\Delta G = - ve$
The relation between Gibbs free energy change and emf of the cell can be written as,
$\Delta G = - nFE$
Where,
$\Delta G$ -Gibbs free energy change
n-number of electrons involved
F-Faraday’s constant
E-emf of the cell
Thus, from the above equation, it is clear that the reaction becomes spontaneous only if the emf of the cell is positive.