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Question: A cell containing two H electrodes. The negative electrode is in contact with a solution of \( {10^{...

A cell containing two H electrodes. The negative electrode is in contact with a solution of 10−6M {10^{ - 6}}M H+{H^ + } ion. The e.m.f of the cell is 0.1180.118 volt at 25∘C{25^\circ }C . [H+]\left[ {{H^ + }} \right] at the positive electrode is :
A. 10−4{10^{ - 4}}
B. 10−5{10^{ - 5}}
C. 10−3{10^{ - 3}}
D. None of the above

Explanation

Solution

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,H2(10−6M)/H+//H+/H2(x)PtPt,{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∘−RTnFlog⁡K......(1)E = {E^\circ } - \dfrac{{RT}}{{nF}}\log K......(1)
Where,
E-emf of the cell
E∘{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∘=0{E^\circ } = 0
The electrons involved between two hydrogen electrodes, n=2n = 2
Substituting all the constant values such as F,R and thus, the term RTnF\dfrac{{RT}}{{nF}} can be written as
RTnF=0.02955\dfrac{{RT}}{{nF}} = 0.02955
(1)→0.118V=−0.05912×log⁡((x)2(10−6)2)(1) \to 0.118V = - \dfrac{{0.0591}}{2} \times \log \left( {\dfrac{{{{(x)}^2}}}{{{{({{10}^{ - 6}})}^2}}}} \right)
⇒0.118=0.05911log⁡(x10−6)\Rightarrow 0.118 = \dfrac{{0.0591}}{1}\log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)
⇒0.05910.118=log⁡(x10−6)\Rightarrow \dfrac{{0.0591}}{{0.118}} = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)
⇒0.1180.0591=log⁡(x10−6)\Rightarrow \dfrac{{0.118}}{{0.0591}} = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)
⇒1.997=log⁡(x10−6)\Rightarrow 1.997 = \log \left( {\dfrac{x}{{{{10}^{ - 6}}}}} \right)
⇒99.31=x10−6\Rightarrow 99.31 = \dfrac{x}{{{{10}^{ - 6}}}}
⇒99.31×10−6=x\Rightarrow 99.31 \times {10^{ - 6}} = x
Since 99.31≈10099.31 \approx 100 , then
⇒10−4=x\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, ΔG=−ve\Delta G = - ve
The relation between Gibbs free energy change and emf of the cell can be written as,
ΔG=−nFE\Delta G = - nFE
Where,
ΔG\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.