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Question: Calculate the electrode potential at \({25^0}{\text{C}}\)of \(C{r^{3 + }}\), \(C{r_2}O_7^{2 - }\) el...

Calculate the electrode potential at 250C{25^0}{\text{C}}of Cr3+C{r^{3 + }}, Cr2O72−C{r_2}O_7^{2 - } electrode at pOH=11pOH = 11 in a solution of 0.01M0.01{\text{M}} both Cr3+C{r^{3 + }} and Cr2O72−C{r_2}O_7^{2 - }
Cr2O72−+14H++6e→2Cr3++7H2OC{r_2}O_7^{2 - } + 14{H^ + } + 6e \to 2C{r^{3 + }} + 7{H_2}O
E0=1.33V{E^0} = 1.33V

Explanation

Solution

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→C+DA + B \to C + D
The Nernst equation is written as
E=E0−RTnFln⁡[A][B][C][D]E = {E^0} - \dfrac{{RT}}{{nF}}\ln \dfrac{{\left[ A \right]\left[ B \right]}}{{\left[ C \right]\left[ D \right]}}
Where, EE denotes the EMF of the electrochemical cell
E0{E^0} denotes the standard cell potential of the redox reaction
nn denotes the number of electrons transferred during the redox reaction
FF denotes Faraday constant
RR denotes the gas constant
TT denotes the temperature of the reaction

Complete step by step answer:
The given cell reaction in the question is Cr2O72−+14H++6e→2Cr3++7H2OC{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 [Cr2O72−]=[Cr3+]=0.01M\left[ {C{r_2}O_7^{2 - }} \right] = \left[ {C{r^{3 + }}} \right] = 0.01{\text{M}}
We are also given pOH=11pOH = 11 and we know that pH+pOH=14pH + pOH = 14
So, we get, pH=3pH = 3 which means [H+]=10−3M\left[ {{H^ + }} \right] = {10^{ - 3}}{\text{M}}
Now using the Nernst Equation for the reaction and substituting the values, we get,
E=1.33−0.0596ln⁡[Cr3+]2[Cr2O72−][H+]14E = 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:
⇒E=1.33−0.0596ln⁡(0.01)2(0.01)(10−3)14\Rightarrow E = 1.33 - \dfrac{{0.059}}{6}\ln \dfrac{{{{\left( {0.01} \right)}^2}}}{{\left( {0.01} \right){{\left( {{{10}^{ - 3}}} \right)}^{14}}}}

∴E=0.936 Volts\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.