Nernst Equation and Its Application

Given that the standard electrode potentials  $(\mathrm{E}°)$ of metals are:

 ${\mathrm{K}}^{+}/\mathrm{K}=-2.93\mathrm{\hspace{0.17em}}\mathrm{V},\mathrm{\hspace{0.17em}}{\mathrm{Ag}}^{+}/\mathrm{Ag}=0.80\mathrm{\hspace{0.17em}}\mathrm{V},\mathrm{\hspace{0.17em}}{\mathrm{Cu}}^{2+}/\mathrm{Cu}=0.34\mathrm{\hspace{0.17em}}\mathrm{V},$

 ${\mathrm{Mg}}^{2+}/\mathrm{Mg}=-2.37\mathrm{\hspace{0.17em}}\mathrm{V},\mathrm{\hspace{0.17em}}{\mathrm{Cr}}^{3+}/\mathrm{Cr}=-0.74\mathrm{\hspace{0.17em}}\mathrm{V},\mathrm{\hspace{0.17em}}{\mathrm{Fe}}^{2+}/\mathrm{Fe}=-0.44\mathrm{\hspace{0.17em}}\mathrm{V}.$

The increasing order of their reducing power is:

Calculate the standard electrode potential of ${\mathrm{Ni}}^{2+}/\mathrm{Ni}$ electrode, if emf of the cell ${\text{Ni\hspace{0.17em}(s)|Ni}}^{2+}\mathrm{\hspace{0.17em}}(0.01\mathrm{\hspace{0.17em}}\mathrm{M})\left|\mathrm{\hspace{0.17em}}\right|\mathrm{\hspace{0.17em}}{\mathrm{Cu}}^{2+}\mathrm{\hspace{0.17em}}(0.1\mathrm{\hspace{0.17em}}\mathrm{M})\left|\mathrm{Cu}\mathrm{\hspace{0.17em}}\right(\mathrm{s})$ is $0.059 \mathrm{V}$.

 ${\text{[Given:\hspace{0.17em}E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=+0.34\mathrm{\hspace{0.17em}}\mathrm{V}]$

The cell $\mathrm{emf}$ and ${\mathrm{\Delta }}_{\mathrm{r}}\mathrm{G}°$ for a cell reaction at $25°\mathrm{C}$ are,

 $\mathrm{Zn}\left(\mathrm{s}\right)\left|{\mathrm{Zn}}^{2+}\right(0.1\mathrm{M}\left)\right|\left|{\mathrm{Cd}}^{2+}\right(0.01\mathrm{M}\left)\right|\mathrm{Cd}\text{\hspace{0.17em}}\left(\mathrm{s}\right)$

 $[\mathrm{Given},{\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\mathrm{o}}=-0.763\mathrm{V},{\mathrm{E}}_{{\mathrm{Cd}}^{2+}/\mathrm{Cd}}^{\mathrm{o}}=-0.403\mathrm{V}]$

 $1\text{\hspace{0.17em}}\mathrm{F}=96,500\mathrm{C}{\mathrm{mol}}^{-1},\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$.

The cell emf and  ${\Delta }_{r}G°$  for the cell reaction at  $\mathrm{}25°C$  are

 $Zn(s)|Z{n}^{2+}(0.1M)||C{d}^{2+}(0.01M)|Cd\text{}(s)$

 $[Given:E_{Z{n}^{2+}/Zn}^o=-0.763V,E_{C{d}^{2+}/Cd}^o=-0.403V]$

 $1\text{}F=96,500Cmo{l}^{-1},R=8.314J{K}^{-1}mo{l}^{-1}]$

The standard reduction potential of  $C{u}^{2+}/CuandA{g}^{+}/Ag$  electrodes is 0.337 and 0.799 volt respectively. Construct a galvanic cell using these electrodes so that its standard e.m.f. is positive. For what concentration of  $A{g}^{+}$will the e.m.f. of the cell at  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C$ be zero if the concentration of  $C{u}^{2+}$  is 0.01 M?

The standard reduction potential for ${\mathrm{Cu}}^{2+}/\mathrm{Cu}$ is $+0.34\mathrm{V}$.  Calculate the reduction potential at  $\mathrm{pH}=14$ for the above couple.  ${\mathrm{K}}_{\mathrm{sp}}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\hspace{0.17em}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\hspace{0.17em}1.0\times {10}^{-19}$.

Two student use same stock solution of ${\mathrm{ZnSO}}_4$ and a solution of ${\mathrm{CuSO}}_4$. The EMF of one cell is 0.03 V higher than the other. The concentration of  ${\mathrm{CuSO}}_4$ in the cell with higher EMF value is 0.5 M. Find out the concentration of ${\mathrm{CuSO}}_4$ in the other cell  $\left(\frac{2\mathrm{.}303\mathrm{RT}}{\mathrm{F}}=0\mathrm{.}06\right)$:

Find the equilibrium constant for the reaction,  $2{\text{Fe}}_{\text{aq}}^{3+}+3{\text{I}}_{\text{aq}}^{-}\rightleftharpoons 2{\text{Fe}}_{\text{aq}}^{2+}+{\text{I}}_{\text{3(aq)}}^{-}$ .  For  ${\text{Fe}}^{3+}/{\text{Fe}}^{2+}{\text{and I}}_3^{-1}/{\text{I}}^{-1}$, the standard reduction potentials in acidic conditions are 0.77V and 0.54V respectively.

An excess of liquid mercury is added to an acidified solution of  $1.0\times {10}^{-3}MF{e}^{3+}.$  It is found that 5% of  $F{e}^{3+}$  remains at equilibrium at  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C$ . Calculate  ${\text{E}}_{\text{H}{\text{g}}_2^{2+}\text{|}\text{Hg}\left(\text{e}\right)}^{\text{o}}$ , assuming that the only reaction that occurs is

 $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}2Hg+\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}2F{e}^{3+}\underset{}{\overset{}{\to }}H{g}_2^{2+}+\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}2F{e}^{2+}.$

(Given  $E_{F{e}^{3+}|F{e}^{2+}}^o=0.77V.)$

Of the halide ions, the most powerful reducing agent is: