Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of \[{10^{ - 6}}{\text{M}}\] hydrogen ions. The EMF of the cell is $0.118{\text{V}}$ at $25^\circ {\text{C}}$ . The concentration of hydrogen ions at the positive electrode is:
A.\[{10^{ - 4}}{\text{M}}\]
B.\[{10^{ - 3}}{\text{M}}\]
C.\[{10^{ - 1}}{\text{M}}\]
D.None of these

Answer 1

Hint: The electrical potential difference set up between the metal and its ions in the solution is referred to as the electrode potential of the cell. Or, in other words, the electrode potential can be said to be the tendency of an electrode to lose electrons or gain electrons when it is in contact with a solution of its own ions.
If oxidation takes place at the electrode with respect to standard hydrogen electrode, then the electrode potential is termed as ‘oxidation potential’. If reduction takes place at the electrode with respect to standard hydrogen electrode, then the electrode potential is termed as ‘reduction potential’.
EMF represents the difference between the electrode potentials of the two half cells when there is no flow of current.

Complete step by step answer:
Given that the negative electrode is in contact with a solution of \[{10^{ - 6}}{\text{M}}\] hydrogen ions. Also given that the EMF of the cell is $0.118{\text{V}}$ at $25^\circ {\text{C}}$ .
We need to find out the concentration of hydrogen ions at the positive electrode.
In an electrochemical cell, the electrode at which oxidation takes place is the anode or the negative electrode. So, we have
At anode, \[{{\text{H}}_{\text{2}}} \to 2{{\text{H}}^{\text{ + }}} + 2{{\text{e}}^{\text{ - }}}\]
The concentration of hydrogen ions at anode ${\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{anode}}}} = {10^{ - 6}}{\text{M}}$
On the other hand, the electrode at which reduction takes place is the cathode or the positive electrode. So, we have
At cathode, \[2{{\text{H}}^{\text{ + }}} + 2{{\text{e}}^{\text{ - }}} \to {{\text{H}}_{\text{2}}}\]
We are to find out the ${\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}}$ .
For a general cell reaction ${\text{aA + bB}}\overset {} \leftrightarrows {\text{cC + dD}}$
The Nernst equation is
\[{{\text{E}}_{{\text{cell}}}} = {{\text{E}}^{\text{0}}}_{{\text{cell}}} - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}\operatorname{l} {\text{og}}\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}\]
Here, n is the number of electrons involved which for the given question is 2.
Also ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ for the hydrogen electrode is 0.
So, we have
${{\text{E}}_{{\text{cell}}}} = - \dfrac{{0.059}}{2}\operatorname{l} {\text{og}}\dfrac{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{anode}}}}^2}}{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{cathode}}}}^2}}$
Substitute all the values.
$
  0.118 = - \dfrac{{0.059}}{2}\operatorname{l} {\text{og}}\dfrac{{{{\left[ {{\text{1}}{{\text{0}}^{{\text{ - 6}}}}} \right]}^{\text{2}}}}}{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{cathode}}}}^{\text{2}}}} \\
   \Rightarrow 0.118 = \dfrac{{0.059}}{2}\operatorname{l} {\text{og}}\dfrac{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{cathode}}}}^{\text{2}}}}{{{{\left[ {{\text{1}}{{\text{0}}^{{\text{ - 6}}}}} \right]}^{\text{2}}}}} \\
   \Rightarrow 0.118 = 0.059\operatorname{l} {\text{og}}\dfrac{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{cathode}}}}}}{{\left[ {{\text{1}}{{\text{0}}^{{\text{ - 6}}}}} \right]}} \\
   \Rightarrow 0.118 = 0.059\left( {\log {{\left[ {{{\text{H}}^{\text{ + }}}} \right]}_{{\text{cathode}}}} - \log \left[ {{\text{1}}{{\text{0}}^{{\text{ - 6}}}}} \right]} \right) \\
   \Rightarrow 0.118 = 0.059 \times \log {\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}} + 0.059 \times 6 \\
   \Rightarrow 0.059 \times \log {\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}} = - 0.236 \\
   \Rightarrow \log {\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}} = - \dfrac{{0.236}}{{0.059}} \\
   \Rightarrow {\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}} = {\text{antilog}}\left( { - 4} \right) \\
   \Rightarrow {\left[ {{{\text{H}}^{\text{ + }}}} \right]_{{\text{cathode}}}} = {10^{ - 4}} \\
 $

So, the correct answer is \[{10^{ - 4}}{\text{M}}\] which is given by option A.

Note:
The cell potential of an electrochemical cell can also be utilized to determine the entropy change for the reaction. Since, entropy is related to the Gibbs free energy through the temperature dependence of the Gibbs free energy by the relation:
${\left( {\dfrac{{\partial \Delta G}}{{\partial T}}} \right)_P} = - \Delta S$