Ramendra C Mukerjee Solutions for Chapter: Electromotive Force, Exercise 1: PROBLEMS

$\mathrm{E}°,$ for the half reactions are as

$$\begin{gathered} \mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\hspace{0.17em};\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=0.76\hspace{0.17em}\mathrm{V} \\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\hspace{0.17em};\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=0.41\hspace{0.17em}\hspace{0.17em}\mathrm{V} \end{gathered}$$

The standard EMF for the cell reaction is:

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$

If $\mathrm{E}{º}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}=-0.44 \mathrm{V}, \mathrm{E}{º}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77 \mathrm{V}$ calculate $\mathrm{E}{º}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}$

Predict reaction of $1 \mathrm{M} \mathrm{HCl}$ with iron. Given, ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{°}=-0.44 \mathrm{V}$.

A cell contains $0.04 \mathrm{M} {\mathrm{Cr}}^{3+}$ in one compartment and $1.0 \mathrm{M} {\mathrm{Cr}}^{3+}$ in the other, with $\mathrm{Cr}$ electrodes in both. Which is the anode compartment?

Under standard conditions for all concentrations, the following reaction is spontaneous at $25°\mathrm{C}$

${\mathrm{O}}_2 \left(\mathrm{g}\right)+4{\mathrm{H}}^{+} \left(\mathrm{aq}\right)+4{\mathrm{Br}}^{-} \left(\mathrm{aq}\right)\to 2{\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right)+2{\mathrm{Br}}_2 \left(\mathrm{l}\right)$

If $\left[{\mathrm{H}}^{+}\right]$ is decreased so, that the $\mathrm{pH}=3.6,$ what value will ${\mathrm{E}}_{\text{cell }}$ have, and will the reaction be spontaneous at this $\left[{\mathrm{H}}^{+}\right]$?

Given: $\begin{array}{ll}{\mathrm{O}}_2+4{\mathrm{H}}^{+}+4\mathrm{e}=2{\mathrm{H}}_2\mathrm{O};& {\mathrm{E}}^0=1·23 \mathrm{V}\\ {\mathrm{Br}}_2+2\mathrm{e}=2{\mathrm{Br}}^{-};& {\mathrm{E}}^0=1·07 \mathrm{V}.\end{array}$

An electrode is prepared by dipping a silver strip into a solution saturated with silver thiocyanate, $\mathrm{AgSCN}$, and containing $0.10 \mathrm{M} {\mathrm{SCN}}^{-}$. The $\mathrm{emf}$ of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half cell as the anode is $0.45 \mathrm{V}$. What is the solubility product of$\mathrm{AgSCN}?$ ${{\mathrm{E}}^0}_{{\mathrm{Ag}}^{+}|\mathrm{Ag}} = 0.80$. If your answer is $\mathrm{y} \times {10}^{-7}. \mathrm{then} \mathrm{y} \mathrm{is}$

Calculate equilibrium constant for the equilibrium,

$2{\mathrm{MnO}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{CO}}_2$

Given that, ${\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/ {\mathrm{Mn}}^{2+}}^0=1.51 \mathrm{V}$ and ${{\mathrm{E}}^0}_{{\mathrm{CO}}_2^{}/{\mathrm{C}}_2{\mathrm{O}}_4^{2-}}=-0.49 \mathrm{V}$. The value of equilibrium constant is ${10}^{\mathrm{x}}$, find the value of $\mathrm{x}$ upto the nearest integer.