Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 2: Level 2

The concentration of ${\mathrm{K}}^{+}$ inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal $\mathrm{M}$ is:

$\mathrm{M}\left(\mathrm{s}\right)\left|{\mathrm{M}}^{+}(\mathrm{aq};0.05\mathrm{M})\Vert {\mathrm{M}}^{+}(\mathrm{aq},1\mathrm{M})\right|\mathrm{M}\left(\mathrm{s}\right)$ 

For the above galvanic cell, the magnitude of the cell potential $\left|E_{\mathrm{cell}}\right|=70 \mathrm{mV}$.

If the $0.05 \mathrm{M}$ solution of ${\mathrm{M}}^{+}$ is replaced by $0.0025 \mathrm{M} {\mathrm{M}}^{+}$ solution, the magnitude of the cell potential would be:

[Hint: From Q. No. 408: $\frac{2.303\mathrm{RT}}{\mathrm{F}}\log \frac{1}{0.05}=0.07;\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.0538.$

Then, apply the Nernst equation again.]

For the reduction of ${\mathrm{NO}}_3^{-}$ ion in an aqueous solution, $\mathrm{E}°$ is $+0.96 \mathrm{V}$. Values of $\mathrm{E}°$ for some metal ions are given below:

$\begin{array}{ll}{\mathrm{V}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{V};& \mathrm{E}°=-1.19 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\to \mathrm{Fe};& \mathrm{E}°=-0.04 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Au}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\to \mathrm{Au};& \mathrm{E}°=+1.40 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Hg}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Hg};& \mathrm{E}°=+0.86 \mathrm{V}\end{array}$
The pair(s) of metals that is (are) oxidised by ${\mathrm{NO}}_3^{-}$ in aqueous solution is (are):

At equimolar concentrations of ${\mathrm{Fe}}^{2+}$ and ${\mathrm{Fe}}^{3+}$, what must $\left[{\mathrm{Ag}}^{+}\right]$ be so that the voltage of the galvanic cell from the $\left({\mathrm{Ag}}^{+}\mid \mathrm{Ag}\right)$ and $\left({\mathrm{Fe}}^{3+}\mid {\mathrm{Fe}}^{2+}\right)$ electrodes equals zero?

${\mathrm{Fe}}^{2+}+{\mathrm{Ag}}^{+}\rightleftharpoons {\mathrm{Fe}}^{3+}+\mathrm{Ag}$,

${\mathrm{E}}_{{\mathrm{Ag}}^{+}\mid \mathrm{Ag}}=0.7991; {\mathrm{E}}_{{\mathrm{Fe}}^{3+}\mid {\mathrm{Fe}}^{2+}}=0.771$

The solubility product of silver iodide is $8.3\times {10}^{-17}$ and the standard reduction potential of $\mathrm{Ag}, {\mathrm{Ag}}^{+}$ electrode is $+0.8 \mathrm{volts}$ at $25 °\mathrm{C}$. The standard reduction potential of $\mathrm{Ag},\mathrm{AgI}|{\mathrm{I}}^{-}$ electrode from these data is:

The standard free energy change (in $\mathrm{J}$) for the reaction $3{\mathrm{Fe}}^{2+ }\left(\mathrm{aqeous}\right)+2\mathrm{Cr} \left(\mathrm{s}\right)\to 2{\mathrm{Cr}}^{3+ }\left(\mathrm{aqeous}\right)+3\mathrm{Fe} \left(\mathrm{s}\right)$ given ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=-0.44 \mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^0=-0.74 \mathrm{V}$ is $\left(\mathrm{F}=96500 \mathrm{C}\right)$.

The standard Gibbs free energy change $\left({\mathrm{\Delta G}}^0 \mathrm{in} \mathrm{kJ} {\mathrm{mol}}^{-1}\right)$ in a Daniel cell $\left(\mathrm{E}°{}_{\mathrm{cell}}=1.1 \mathrm{V}\right)$, when $2$ moles of $\mathrm{Zn}\left(\mathrm{s}\right)$ is oxidised at $298 \mathrm{K}$, is closed to:

An electrochemical cell is made by placing a zinc electrode in $1.0 \mathrm{litre}$of $0.2 \mathrm{M}$ ${\mathrm{ZnSO}}_4$ solution and a copper electrode in $1.0 \mathrm{litre}$ of $0.015 \mathrm{M} {\mathrm{CuCl}}_2$ solution. What is the initial voltage of this cell when it is properly constructed?

${\mathrm{E}}_{\mathrm{cell}}^0 = 1.1 \mathrm{V}$

Answer correct up to two places of decimals.

Calculate the ratio of the reduced to the oxidised form at half-cell potential of $0.1 \mathrm{V}$, for the half cell ${\mathrm{Fe}}^{3+}, {\mathrm{Fe}}^{2+}|\mathrm{Pt}$. The ratio is in the form of $\mathrm{x}\times {10}^{11}$, give the value of $\mathrm{x}$.

${{\mathrm{E}}^{\circ }}_{{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}}=0.7591 \mathrm{V}, \frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{antilog} 0.17=1.48$.