Question

Calculate the electrode potential at ${25^0}{\text{C}}$of $C{r^{3 + }}$, $C{r_2}O_7^{2 - }$ electrode at $pOH = 11$ in a solution of $0.01{\text{M}}$ both $C{r^{3 + }}$ and $C{r_2}O_7^{2 - }$
$C{r_2}O_7^{2 - } + 14{H^ + } + 6e \to 2C{r^{3 + }} + 7{H_2}O$
${E^0} = 1.33V$

Answer 1

Hint:To answer this question, you must recall the Nernst Equation. Nernst equation gives a relation between the EMF, temperature and the concentrations of chemical species of a redox reaction.
Formula used: For a reaction, $A + B \to C + D$
The Nernst equation is written as
$E = {E^0} - \dfrac{{RT}}{{nF}}\ln \dfrac{{\left[ A \right]\left[ B \right]}}{{\left[ C \right]\left[ D \right]}}$
Where, $E$ denotes the EMF of the electrochemical cell
${E^0}$ denotes the standard cell potential of the redox reaction
$n$ denotes the number of electrons transferred during the redox reaction
$F$ denotes Faraday constant
$R$ denotes the gas constant
$T$ denotes the temperature of the reaction

Complete step by step answer:
The given cell reaction in the question is $C{r_2}O_7^{2 - } + 14{H^ + } + 6e \to 2C{r^{3 + }} + 7{H_2}O$
We are given the concentrations of the ions in the solution as $\left[ {C{r_2}O_7^{2 - }} \right] = \left[ {C{r^{3 + }}} \right] = 0.01{\text{M}}$
We are also given $pOH = 11$ and we know that $pH + pOH = 14$
So, we get, $pH = 3$ which means $\left[ {{H^ + }} \right] = {10^{ - 3}}{\text{M}}$
Now using the Nernst Equation for the reaction and substituting the values, we get,
$E = 1.33 - \dfrac{{0.059}}{6}\ln \dfrac{{{{\left[ {C{r^{3 + }}} \right]}^2}}}{{\left[ {C{r_2}O_7^{2 - }} \right]{{\left[ {{H^ + }} \right]}^{14}}}}$
Substituting the values:
$ \Rightarrow E = 1.33 - \dfrac{{0.059}}{6}\ln \dfrac{{{{\left( {0.01} \right)}^2}}}{{\left( {0.01} \right){{\left( {{{10}^{ - 3}}} \right)}^{14}}}}$

$\therefore E = 0.936{\text{ Volts}}$

Note:
The Nernst equation helps to calculate the extent of reaction occurring between two redox systems and is thus, generally used to determine if a particular reaction would go to completion or not. At equilibrium, the EMFs of the two half cells are equal. This enables us to calculate the equilibrium constant and hence, the extent of the reaction.
Limitations of Nernst Equation: Nernst equation can be expressed directly in the terms of concentrations of constituents in dilute solutions. But at higher concentrations, the true activities of the ions become significant and therefore, must be used. This complicates the Nernst equation, as estimation of these non-ideal activities of ions requires complex experimental measurements. Also, the Nernst equation applies only when there is no net current flow through the electrode.