Question

Electrolysis of a solution of $MnS{O_4}$ in aqueous sulphuric acid is a method for the preparation of $Mn{O_2}$ as per the reaction

\[{{M}}{{{n}}^{2 + }}(aq.) + 2{H_2}O \to Mn{O_2} + 2{H^ + }(aq.) + {H_2}\]

Passing a current of \[27A\] for \[24\] hours gives one kg of \[Mn{O_2}\]. What is the value of current efficiency? Write the reactions taking place at the cathode and at the anode.

Answer 1

Hint: According to Faraday’s law of electrolysis, the weight of a chemical substance deposited at an electrode after electrolysis is equal to the product of the equivalent weight of the substance, the current passing through the electrodes in amperes, and the time, divided by 96500 coulombs (the value of 1 Faraday).

Complete step by step answer:
The weight of a substance deposited at a cathode or anode is given by Faraday’s law of electrolysis in the form of the given equation:
\[Weight\;in\;grams = \dfrac{{Equivalent\;weight \times Current\;in\;amperes \times Time\;in\;seconds}}{{96500}}\]
\[Equivalent\;weight = \dfrac{{Molecular\;weight}}{{Change\;in\;oxidation\;number\;}}\]
Molecular weight of \[Mn{O_2} = Molecular\;weight\;of\;Mn + \left( {2 \times Molecular\;weight\;of\;O} \right)\]
\[ \Rightarrow \] Molecular weight of \[Mn{O_2} = 56g + \left( {2 \times 16g} \right)\]
\[ \Rightarrow \] Molecular weight of \[Mn{O_2} = 56g + 32g = 87g\]
The oxidation number of \[Mn\] in \[{{M}}{{{n}}^{2 + }} = + 2\]
The oxidation number of \[Mn\] in \[Mn{O_2}\] can be calculated as follows:
\[x + \left( {2 \times - 2} \right) = 0\]
\[ \Rightarrow x - 4 = 0\]
\[ \Rightarrow x = + 4\]
Therefore, change in oxidation number \[ = + 4 - \left( { + 2} \right) = + 4 - 2 = 2\]
Hence, the equivalent weight of \[Mn{O_2} = \dfrac{{87g}}{2} = 43.g\]
Given,
Weight of \[Mn{O_2} = 1kg = 1000g\]
Current \[ = 27A\]
Time $ = 24hours$
According to the equation, the time must be in seconds, we have to convert 24 hours into seconds.
Now, we know that, \[{{1hour = 60minutes}}\], and \[{{1minute = 60seconds}}\]
So, to convert hour into seconds, we shall first change hours into minutes by multiplying it by $60$ and then further converting minutes into seconds by multiplying it by $60$.
Therefore, \[24\;hrs.\; = (24\; \times 60 \times 60)\;sec\]
Substituting all the given values in Faraday’s equation, we get;
Or, \[1000\;g\; = \dfrac{{43.5g \times i \times (24 \times 60 \times 60)\;sec}}{{96500}}\] \[ \Rightarrow \] \[i = 25.67A\]

Current efficiency can be defined as the amount (percentage) of current passing through the electrode in comparison to the actual current being applied across the electrode during electrolysis of a solution.
Applying this definition, the value of current efficiency during the electrolysis of \[MnS{O_4}\] can be calculated as:
$CurrentEfficiency = \dfrac{{25.67}}{{27}} \times 100\% $
\[ \Rightarrow \;Current\;Efficiency\; = \;95.074\;\% \]
Therefore, the value of the current efficiency is $95.074\% $.
The reaction at the cathode is a reduction reaction in which one or more electrons are gained by an ion.
The following reaction takes place at the cathode:
 \[2{H^ + } + 2{e^ - } \to {H_2}\]
The reaction at the anode is an oxidation reaction in which one or more electrons are lost by an ion.
The following reaction takes place at the anode:
\[{{M}}{{{n}}^{2 + }} \to {{M}}{{{n}}^{4 + }} + 2{e^ - }\]

Note: Reduction reaction at the cathode:
The cathode is the electrode that is negatively charged and has an excess of electrons.
The ion which moves towards the cathode gains electrons from it and gets reduced to a neutral molecule.

Oxidation reaction at the anode:
The anode is the electrode that is positively charged and has a lack of electrons.
The ion which moves towards the anode loses its electrons to it and gets oxidized to a more positively charged cation.