Question

Calculate the electrode potential at ${25^ \circ }{\text{C}}$ of ${\text{C}}{{\text{r}}^{3 + }}$, ${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }$ electrode at ${\text{pOH}} = 11$ in a solution of $0.01{\text{ M}}$ both in ${\text{C}}{{\text{r}}^{3 + }}$ and ${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }$.
${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - } + 14{{\text{H}}^ + } + 6{\text{e}} \to 2{\text{C}}{{\text{r}}^{3 + }} + 7{{\text{H}}_2}{\text{O }}{{\text{E}}^0} = 1.33{\text{ V}}$

Answer 1

Hint: The difference in the potential of two half cells is known as the electrode potential. Electrode potential is denoted by E. when the concentration of all the reactant and product species in the reaction is unity is known as the standard electrode potential. Standard electrode potential is denoted by ${{\text{E}}^0}$.

Complete step by step answer:
We are given that the pOH of the solution is 11. From the pOH calculate the pH of the solution using the equation as follows:
${\text{pH}} + {\text{pOH}} = 14$
Rearrange the equation for pH. Thus,
${\text{pH}} = 14 - {\text{pOH}}$
Substitute 11 for pOH. Thus,
${\text{pH}} = 14 - {\text{11}}$
${\text{pH}} = 3$
The pH of the solution is the negative logarithm of the hydrogen ion concentration. Thus,
$\left[ {{{\text{H}}^ + }} \right] = {10^{ - {\text{pH}}}}$
$\left[ {{{\text{H}}^ + }} \right] = {10^{ - 3}}{\text{ M}}$
Thus, the concentration of the hydrogen ion in the solution is ${10^{ - 3}}{\text{ M}}$.
The given reaction is,
${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - } + 14{{\text{H}}^ + } + 6{\text{e}} \to 2{\text{C}}{{\text{r}}^{3 + }} + 7{{\text{H}}_2}{\text{O}}$
From the reaction, we can conclude that the number of moles of electrons involved in the reaction are $6{\text{ mol}}$.
Calculate the electrode potential using the equation as follows:
${\text{E}} = {{\text{E}}^0} - \dfrac{{0.0591}}{n}\log \dfrac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactant}}} \right]}}$
Where ${\text{E}}$ is the electrode potential,
${{\text{E}}^0}$ is the standard electrode potential,
$n$ is the number of moles of electrons.
From the given reaction, we can write the equation for electrode potential as,
${\text{E}} = {{\text{E}}^0} - \dfrac{{0.0591}}{n}\log \left( {\dfrac{{{{\left[ {{\text{C}}{{\text{r}}^{3 + }}} \right]}^2}}}{{\left[ {{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }} \right]{{\left[ {{{\text{H}}^ + }} \right]}^{14}}}}} \right)$
Substitute $1.33{\text{ V}}$ for the standard electrode potential, $6{\text{ mol}}$ for the number of moles of electrons, $0.01{\text{ M}}$ for the concentration of ${\text{C}}{{\text{r}}^{3 + }}$ and ${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }$, ${10^{ - 3}}{\text{ M}}$ for the concentration of ${{\text{H}}^ + }$ ion. Thus,
${\text{E}} = 1.33{\text{ V}} - \dfrac{{0.0591}}{{6{\text{ mol}}}}\log \left( {\dfrac{{{{\left[ {0.01{\text{ M}}} \right]}^2}}}{{\left[ {0.01{\text{ M}}} \right]{{\left[ {{{10}^{ - 3}}{\text{ M}}} \right]}^{14}}}}} \right)$
${\text{E}} = 1.33{\text{ V}} - \dfrac{{0.0591}}{{6{\text{ mol}}}}\log \left( {{{10}^{40}}} \right)$
${\text{E}} = 0.936{\text{ V}}$

Thus, the electrode potential is $0.936{\text{ V}}$. .

Note: The potential of a cell measured under standard conditions of temperature and pressure is known as standard electron potential. The standard conditions of temperature and pressure means the temperature is $298{\text{ K}}$ and the pressure is $1{\text{ atm}}$.