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

The equivalent conductance of $0.10 \mathrm{N}$ solution of ${\mathrm{MgCl}}_2$ is $97.1{\mathrm{mhocm}}^2$ ${\mathrm{eq}}^{-1}$. A cell with electrodes that are $1.50{ \mathrm{cm}}^2$ in surface are and $0.50 \mathrm{cm}$ apart is filled with $0.1 {\mathrm{NMgCl}}_2$ solution. How much current will flow when the potential difference between the electrodes is $5$ volts? If your answer is $x\times {10}^{-1}$ then $\mathrm{x}$ is:

Calculate ${\mathrm{K}}_{\mathrm{a}}$ of acetic acid if its $0.05 \mathrm{N}$ solution has equivalent conductance of $7.36 \mathrm{m} \mathrm{ho} {\mathrm{cm}}^2$ at $25°\mathrm{C}$. $\left({\mathrm{\lambda }}_{{\mathrm{CH}}_3\mathrm{COOH}}^{\infty }=390.70\right)$ If your answer is $\mathrm{x}\times {10}^{-5}$ then $\mathrm{x}$ is :

$\mathrm{Cd}$ amalgam is prepared by electrolysis of a solution ${\mathrm{CdCl}}_2$ using a mercury cathode. What time (in sec.) current of $5$ A should be passed in order to prepare $12\%\mathrm{Cd}-\mathrm{Hg}$ amalgam on a cathode of $2 \mathrm{g}, \mathrm{Hg}$ (atomic weight of $\mathrm{Cd}=112.4$)

After electrolysis of $\mathrm{NaCl}$ solution with inert electrodes for a certain period of time, $600 \mathrm{mL}$ of the solution was left. Which was found to be $1 \mathrm{N}$ in $\mathrm{NaOH}$. During the same time, $31.75 \mathrm{g}$ of $\mathrm{Cu}$ deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage yield of $\mathrm{NaOH}$ obtained.

The e.m.f. of the cell obtained by combining $\mathrm{Zn}$ and $\mathrm{Cu}$ electrode of a Daniel cell with $\mathrm{N}$ calomel electrode in two different arrangements are $1.083 \mathrm{V}$ and $0.018 \mathrm{V}$ respectively at $25°\mathrm{C}$. If the standard reduction potential of $\mathrm{N}$ calomel electrode is $0.28 \mathrm{V}$ find the emf of Daniel cell.

A current of $3.7 \mathrm{A}$ is passed for $6$ hrs between $\mathrm{Pt}$ electrodes in $0.5 \mathrm{L}$ of $2 \mathrm{M}$ solution of $\mathrm{Ni}{\left({\mathrm{NO}}_3\right)}_2$. What will be the molarity of solution at the end of electrolysis?

The cell $\mathrm{Pt},{\mathrm{H}}_2(1 \mathrm{atm})\left|{\mathrm{H}}^{+}(\mathrm{pH}=\mathrm{x})\right|\mid$Normal calomel Electrode has an EMF of $0.67 \mathrm{V}$ at $25°\mathrm{C}$. Calculate the $\mathrm{pH}$ of the solution. The oxidation potential of the calomel electrode on hydrogen scale is $-0.28 \mathrm{V}$.

Calculate the voltage $\mathrm{E}$, of the cell at $25°\mathrm{C}$

$\mathrm{Mn}\left(\mathrm{s}\right)\left|\mathrm{Mn}\left({\mathrm{OH}}_2\right)\left(\mathrm{s}\right)\right|{\mathrm{Mn}}^{2+}\left(\mathrm{xM}\right),{\mathrm{OH}}^{-}\left(1.00\times {10}^{-4}\mathrm{M}\right)\left|\right|{\mathrm{Cu}}^{2+}(0.675\mathrm{M})\mid \mathrm{Cu}\left(\mathrm{s}\right)$

given that ${\mathrm{K}}_{\mathrm{sp}}=1.9\times {10}^{-13}$ for $\mathrm{Mn}\left(\mathrm{OH}{)}_2\right(\mathrm{s})$ $\mathrm{E}°\left(\frac{{\mathrm{Mn}}^{2+}}{\mathrm{Mn}}\right)=-1.18 \mathrm{V},\mathrm{E}°\left(\frac{{\mathrm{Cu}}^{+2}}{\mathrm{Cu}}\right)=+0.34 \mathrm{V}$