Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 1: JEE Main - 15th April 2018

The equilibrium constant for the reaction $\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Sn}\left(\mathrm{s}\right)$ is $1\times {10}^{20}$ at $298 \mathrm{K}$. The magnitude of standard electrode potential of $\mathrm{Sn}/{\mathrm{Sn}}^{2+}$ if ${\mathrm{E}}_{{\mathrm{Zn}}^{2*}/\mathrm{Zn}}^0=-0.76 \mathrm{V}$ is _____ $\times {10}^{-2} \mathrm{V}$. (Nearest integer)

Given : $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V}$

Consider the cell

${\mathrm{Pt}}_{\left(\mathrm{s}\right)}\left|{\mathrm{H}}_2\left(\mathrm{g},1\mathrm{atm}\right)\right|{\mathrm{H}}^{+}\left(\mathrm{aq},1\mathrm{M}\right)|\left|{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right),{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\right|Pt\left(\mathrm{s}\right)$

When the potential of the cell is $0.712 \mathrm{V}$ at $298 \mathrm{K}$, the ratio $\left[{\mathrm{Fe}}^{2+}\right]/\left[{\mathrm{Fe}}^{3+}\right]$ is
(Nearest integer)
Given: $\overline{){\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}={\mathrm{Fe}}^{2+},\mathrm{E}°{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}}\mathrm{Pt}=0.771 \frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{V}$

The electrode potential of the following half cell at $298 \mathrm{K}$

$\mathrm{X}\left|{\mathrm{X}}^{2+}\left(0.001\mathrm{M}\right)\Vert {\mathrm{Y}}^{2+}\left(0.01\mathrm{M}\right)\right|\mathrm{Y}$ is _____ $\times {10}^{-2} \mathrm{V}$ (Nearest integer)

Given : ${\mathrm{E}}_{{\mathrm{X}}^{2+}\mid \mathrm{X}}^0=-2.36 \mathrm{V}$

${\mathrm{E}}_{{\mathrm{Y}}^{2+}\mathrm{Y}}^0=+0.36 \mathrm{V}$

$\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{V}$

Which one of the following statements is correct for electrolysis of brine solution?

The logarithm of equilibrium constant for the reaction ${\mathrm{Pd}}^{2+}+4{\mathrm{Cl}}^{-}\rightleftharpoons {\mathrm{PdCl}}_4^{2-}$ is
(Nearest integer)

Given: $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{V}$

${\mathrm{Pd}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pd}\left(\mathrm{s}\right) {\mathrm{E}}^o=0.83 \mathrm{V}$

${\mathrm{PdCl}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pd}\left(\mathrm{s}\right)+4{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{E}}^{\mathrm{o}}=0.65 \mathrm{V}$

The resistivity of a $0.8\mathrm{M}$ solution of an electrolyte is $5\times {10}^{-3}\mathrm{\Omega cm}$. Its molar conductivity is ${10}^4{\mathrm{\Omega }}^{-1} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$. (Nearest integer)

At what $\mathrm{pH}$, given half cell ${\mathrm{MnO}}_4^{-}(0.1\mathrm{M})\mid {\mathrm{Mn}}^{2+}$ $(0.001 \mathrm{M})$ will have electrode potential of $1.282 \mathrm{V}$ ? (Nearest Integer)

Given ${\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/{\mathrm{Mn}}^{2+}}^o=1.54 \mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V}$

$1\times {10}^{-5} \mathrm{M} {\mathrm{AgNO}}_3$ is added to $1 \mathrm{L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at $298 \mathrm{K}$ is $\_\_\_\_\_\_\_\times {10}^{-8} \mathrm{S} {\mathrm{m}}^{-1}$.

[Given : ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgBr}\right)=4.9\times {10}^{-13}$ at $298 \mathrm{K}$

${\lambda }_{{\mathrm{Ag}}^{+}}^0=6\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}$

${\lambda }_{{\mathrm{Br}}^{-}}^0=8\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}$

$\left.{\lambda }_{{\mathrm{NO}}_3^{-}}^0=7\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}\right]$