Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contains 0.1M $MnO_{4}^{-}$ and 0.8 M ${{H}^{+}}$and which was treated with $F{{e}^{2+}}$ necessary to reduce 90% of $MnO_{4}^{-}$ to $M{{n}^{2+}}$.$E_{{MnO_{4}^{-}}/{M{{n}^{2+}}}\;}^{0}=1.51V$.Give answer in V in nearest integer.

Answer 1

To solve the problems in which the concentrations of the reactants and products taking part in the cell reaction is given , we use the Nernst equation to the potential of the indicator electrode.
And the equation is as follows:
${{\text{E}}_{\text{cell}}}\text{=}{{\text{E}}^{\text{0}}}_{\text{cell}}\,\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{n}}\text{log}\dfrac{\left[ \text{products} \right]}{\left[ \text{reactants} \right]}$

Complete step-by-step answer: So in the question, it is asked that what will be the reduction potential of an indicator electrode versus SHE, if the concentration $MnO_{4}^{-}$ is 0.1M and ${{H}^{+}}$is 0.8M. And the 90% of $MnO_{4}^{-}$ was converted to $M{{n}^{2+}}$
First let us write the overall reaction happening in the cell to do further calculations.
The net reaction is: $MnO_{4}^{-}+5F{{e}^{2+}}+8{{H}^{+}}\to 5F{{e}^{3+}}+M{{n}^{2+}}+4{{H}_{2}}O$
The reduction half reaction is as follows:
$MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$
Let us only take the case of reduction reaction in which $MnO_{4}^{-}$is getting reduced to $M{{n}^{2+}}$and trace the change in the concentration of the reactants and products.While proceeding the reaction 90% of $MnO_{4}^{-}$ is converted to $M{{n}^{2+}}$.

The equation is:	$MnO_{4}^{-}+{{H}^{+}}\to M{{n}^{2+}}$
Initial concentration(M)	0.1
Change in concentration(M)	$0.1-x$
Final concentration(M)	0.01

So now we got the values of the final concentrations of the products and the reactants involved in the reaction.
The product concentration i.e. $\left[ M{{n}^{2+}} \right]\,=\,0.09M$
And the concentration of reactants are $\left[ MnO_{4}^{-} \right]\,=\,0.01M$ and $\left[ {{H}^{+}} \right]=0.08M$
Substitute these value in Nernst equation,
${{\text{E}}_{\text{cell}}}\text{=}{{\text{E}}^{\text{0}}}_{\text{cell}}\,\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{n}}\text{log}\dfrac{\left[ \text{products} \right]}{\left[ \text{reactants} \right]}$
$\Rightarrow E_{cell}^{0}=1.51V$
$\Rightarrow {{E}_{cell}}=1.51-\dfrac{0.0592}{5}\log \dfrac{0.09}{\left( 0.01 \right)\times {{\left( 0.08 \right)}^{8}}}$
$\Rightarrow {{E}_{cell}}\,=\,1.51\,-\,0.099\,=\,1.411V$
And hence the potential of the electrode is 1.41V.

Note: For solving these questions, we should always write the net reactions first and the equation should be a balanced one.
Some students forget to calculate the power terms associated with the concentration terms while doing the calculation part. And caution should be taken while taking the log values, always it is better to divide all the values associated with the log first, then take the logarithmic value of the obtained value.