Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $\text{ 1}{{\text{0}}^{-6}}\text{ M }$ hydrogen ion. The EMF of the cell is $\text{ 0}\text{.118 V }$ at $\text{ 2}{{\text{5}}^{\text{0}}}\text{C }$ . Calculate the concentration of hydrogen ions at the positive electrode.

Answer 1

Suppose an electrode is placed in a different concentration in the same solution ten the general construction cell is represented as,
$\text{ Pt }\\!\\!|\\!\\!\text{ }{{a}_{\text{1}}}\left( 1\text{ atm} \right)|\text{ }\underset{{{\text{C}}_{\text{1}}}\text{M}}{\mathop{{{a}^{\text{+}}}}}\,||\underset{{{\text{C}}_{\text{2}}}\text{M}}{\mathop{{{a}^{\text{+}}}}}\,\text{ }\\!\\!|\\!\\!\text{ }{{a}_{2}}\left( 1\text{ atm} \right)|\text{Pt }$
The EMF of the concentration cell is expressed by the modified Nernst equation. It is shown as follows,
$\text{ }{{\text{E}}_{\text{cell }}}\text{= }\dfrac{0.0591}{\text{n}}\text{ }\log \text{ }\dfrac{\left[ \text{Oxidised state} \right]}{\left[ \text{Reduced state} \right]}\text{ }$
Where n is the number of electrons involved in a redox reaction.

Complete step by step solution:
The hydrogen gas concentration cell can be represented as follows,
$\text{ Pt }\\!\\!|\\!\\!\text{ }{{\text{H}}_{\text{1}}}\left( 1\text{ atm} \right)|\text{ }\underset{{{10}^{-6}}\text{M}}{\mathop{{{\text{H}}^{\text{+}}}}}\,||\underset{\text{CM}}{\mathop{{{\text{H}}^{\text{+}}}}}\,\text{ }\\!\\!|\\!\\!\text{ }{{\text{H}}_{2}}\left( 1\text{ atm} \right)|\text{Pt }$
At the negative electrode that is at the cathode, the hydrogen gas undergoes the reduction. The reduction reaction of the hydrogen gas at the cathode is represented as follows
$\text{ At cathode : }{{\text{H}}_{\text{2}}}\text{ }\to \text{ 2}{{\text{H}}^{\text{+}}}\text{ + 2}{{\text{e}}^{-}}\text{ }$
At the positive electrode that is an anode, the hydrogen ions undergo the oxidation reaction .the oxidation reaction of the hydrogen ions at the anode is represented as follows,$\text{ At anode : 2}{{\text{H}}^{\text{+}}}\text{+ 2}{{\text{e}}^{-}}\text{ }\to \text{ }{{\text{H}}_{\text{2}}}\text{ }$
We are interested in determining the hydrogen ion concentration.
The Nernst equation for the hydrogen electrode is written as follows,
$\text{ }{{\text{E}}_{\text{cell }}}\text{= }\dfrac{0.0591}{\text{n}}\text{ }\log \text{ }\dfrac{\left[ \text{Oxidised state} \right]}{\left[ \text{Reduced state} \right]}\text{ }$
Here, the concentration of hydrogen gas is given as the $\text{ 1}{{\text{0}}^{-6}}\text{ M }$and the redox reaction involves the 2 electrons. Thus the Nernst equation is modified as,
$\text{ }{{\text{E}}_{\text{cell }}}\text{= }\dfrac{0.0591}{2}\text{ }\log \text{ }\dfrac{\text{ }\left[ {{\text{H}}^{+}} \right]_{_{\text{cathode}}}^{2}}{{{\left[ {{10}^{-6}} \right]}^{2}}}\text{ }$
On further solving we have,
$\begin{aligned} & \text{ }{{\text{E}}_{\text{cell }}}\text{= }\dfrac{0.0591}{{}}\text{ }\times \text{}\times \log \text{ }\dfrac{\left[ {{\text{H}}^{+}} \right]}{\left[ {{10}^{-6}} \right]}\text{ } \\\ & \Rightarrow \text{ 0}\text{.118 = }\left( 0.0591 \right)\times \log \text{ }\dfrac{\left[ {{\text{H}}^{+}} \right]}{\left[ {{10}^{-6}} \right]} \\\ & \Rightarrow \text{ }\dfrac{\text{0}\text{.118}}{0.0591}\text{ = 2 = }\log \text{ }\dfrac{\left[ {{\text{H}}^{+}} \right]}{\left[ {{10}^{-6}} \right]}\text{ } \\\ \end{aligned}$
Take an antilog of the logarithmic value .We can get the value of hydrogen ion concentration. We get,
$\begin{aligned} & \text{ }\dfrac{\left[ {{\text{H}}^{+}} \right]}{{{10}^{-6}}}=\text{1}{{\text{0}}^{\text{2}}}\text{ } \\\ & \Rightarrow \left[ {{\text{H}}^{+}} \right]\text{ = 1}{{\text{0}}^{\text{2}}}\times {{10}^{-6}} \\\ & \therefore \text{ }\left[ {{\text{H}}^{+}} \right]\text{ = 1}{{\text{0}}^{-\text{4}}}\text{ M } \\\ \end{aligned}$

Therefore the hydrogen ion concentration at the positive electrode or the oxidation reaction is equal to $\text{ 1}{{\text{0}}^{-\text{4}}}\text{ M }$.

Note: Note that, the concentration cells consist of the anode and the cathode are separated by the invisible membrane these compartments are identical except the concentration of solution