
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
459k+ views
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Who is Mukesh What is his dream Why does it look like class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE
