Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 3: Exercise-3

For the following electrochemical cell at $298 \mathrm{K}$, $\mathrm{Pt}\left(\mathrm{s}\right)\overline{){\mathrm{H}}_2\left(\mathrm{g}\right)(1 \mathrm{bar})\overline{){{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}(1 \mathrm{M})\parallel {{\mathrm{M}}^{4+}}_{\left(\mathrm{aq}\right)},{{\mathrm{M}}^{2+}}_{\left(\mathrm{aq}\right)}\overline{)\mathrm{Pt}\left(\mathrm{s}\right)}}}$, ${\mathrm{E}}_{\mathrm{cell}}=0.092 \mathrm{V}$ when $\frac{\left[{{\mathrm{M}}^{2+}}_{\left(\mathrm{aq}\right)}\right]}{\left[{{\mathrm{M}}^{4+}}_{\left(\mathrm{aq}\right)}\right]}={10}^{\mathrm{x}}$. Given : ${{\mathrm{E}}^{°}}_{({\mathrm{M}}^{4+}/{\mathrm{M}}^{2+})}=0.151 \mathrm{V}; 2.303\frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V}$. The value of $\mathrm{x}$ is

For the following cell, $\mathrm{Zn}\left(\mathrm{s}\right)\overline{){{\mathrm{ZnSO}}_4}_{\left(\mathrm{aq}\right)}\parallel {{\mathrm{CuSO}}_4}_{\left(\mathrm{aq}\right)}\overline{)\mathrm{Cu}\left(\mathrm{s}\right)}}$ when the concentrations of ${\mathrm{Zn}}^{2+}$ is $10$ times the concentration of ${\mathrm{Cu}}^{2+}$, the expression for $∆\mathrm{G}$ (in $\mathrm{J} {\mathrm{mol}}^{-1}$) is $[$$\mathrm{F}$ is Faraday constant; $\mathrm{R}$ is gas constant; $\mathrm{T}$is temperature; ${{\mathrm{E}}^{°}}_{\mathrm{cell}}=1.1 \mathrm{V} ]$

Consider an electrochemical cell: $\mathrm{A}\left(\mathrm{s}\right)\overline{){{\mathrm{A}}^{\mathrm{n}}}^{+}}(\mathrm{aq}, 2\mathrm{M})\parallel {\mathrm{B}}^{2\mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M})\overline{)\mathrm{B}\left(\mathrm{s}\right)}$, the value of $∆\mathrm{H}°$ for the cell reaction is twice that of $∆\mathrm{G}°$ at $300 \mathrm{K}$. If the emf of the cell is zero, $∆\mathrm{S}°$(in ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$) of the cell reaction per mole of  $\mathrm{B}$ formed at $300 \mathrm{K}$ is (Given : $\ln \left(2\right)=0.7, \mathrm{R}=8.31 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}.\mathrm{H},\mathrm{S}$and $\mathrm{G}$ are enthalpy, entropy and Gibbs energy, respectively)

Given ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}^0=0.036 \mathrm{V}, {\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=-0.439 \mathrm{V}$

The value of standard electrode potential for the charge, ${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}$ (aq) will be

The standard reduction potential for ${\mathrm{Zn}}^{\mathrm{2+}}/\mathrm{Zn}, {\mathrm{Ni}}^{\mathrm{2+}}/\mathrm{Ni} \text{and} {\mathrm{Fe}}^{\mathrm{2+}}/\mathrm{Fe}$ are $-0.76 \mathrm{V}, -0.23 \mathrm{V} \mathrm{and} -0.44 \mathrm{V}$, respectively. The reaction $\mathrm{X}+{\mathrm{Y}}^{+2}\stackrel{}{\to } {\mathrm{X}}^{2+}+\mathrm{Y}$ will be spontaneous when ______.

The equivalent conductance of $\mathrm{NaCl}$ at concentration $\mathrm{C}$ and at infinite dilution are ${\mathrm{\lambda }}_{\mathrm{C}}$ and ${\mathrm{\lambda }}_{\infty }$ respectively. The correct relationship between ${\mathrm{\lambda }}_{\mathrm{C}}$ and ${\mathrm{\lambda }}_{\infty }$ is given as

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is?

To find the standard potential of ${\mathrm{M}}^{3+}\mid \mathrm{M}$ electrode, the following cell is constituted : $\mathrm{Pt}\left|\mathrm{M}\right|{\mathrm{M}}^{3+}\left(0.001 \mathrm{mol} {\mathrm{L}}^{-1}\right)\left|\left|{\mathrm{Ag}}^{+}\left(0.01 \mathrm{mol} {\mathrm{L}}^{-1}\right)\right|{\mathrm{Ag}}^{+}\left(0.01 {\mathrm{molL}}^{-1}\right)\right|\mathrm{Ag}$ The emf of the cell is found to be $0.421 \mathrm{volt}$ at $298\mathrm{K}$. The standard potential of half reaction ${\mathrm{M}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{M}$ at $298\mathrm{K}$ will be (Given ${\mathrm{E}}_{{\mathrm{Ag}}^{*}/\mathrm{Ag}}^{\mathrm{O}}$ at $298\mathrm{K}=0.80 \mathrm{volt}$)