Nernst Equation and its Applications

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,

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

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

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

Zinc granules are added in excess to a $500$ mL of $1.0$ M nickel nitrate solution at  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C$  until the equilibrium is reached. If the standard reduction potential of  ${\mathrm{Zn}}^{2+}/\mathrm{Zn}\mathrm{and}{\mathrm{Ni}}^{2+}/\mathrm{Ni}$  are  $-0.75\mathrm{V}\mathrm{and}-0.24\mathrm{V}$  respectively, the concentration of  ${\mathrm{Ni}}^{2+}$  in solution at equilibrium.

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 at  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C$ of the reaction,  $2H_{2}O+2e^{-}\underset{}{\rightleftharpoons }{H}_2+\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}2O{H}^{-}is-0.8277V.$  The equilibrium constant for the reaction  $2H_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}+O{H}^{-}$  at  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C.$

The equilibrium constant for the reaction would be:

${\text{Fe}}_{\left(a\text{q}\right)}^{2+}+{\text{Ce}}_{\left(a\text{q}\right)}^{3+}\rightleftharpoons {\text{Ce}}_{\left(a\text{q}\right)}^{2+}+{\text{Fe}}_{\left(a\text{q}\right)}^{3+}$ Given that ${\mathrm{E}}_{{\text{Ce}}^{3+}/{\text{Ce}}^{2+}}^{\text{o}}=1\text{.}44\text{V,}{ \mathrm{E}}_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}^{\text{o}}=0\text{.}68\text{V}$

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.  ${\mathrm{K}}_{\mathrm{sp}}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\hspace{0.17em}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\hspace{0.17em}1.0\times {10}^{-19}$.

Find the equilibrium constant for the reaction,

 ${\text{Cu}}^{2+}+{\text{In}}^{2+}\rightleftharpoons {\text{Cu}}^{+}+{\text{In}}^{3+}$

Given  ${\mathrm{E}}_{{\text{Cu}}^{2+}/{\text{Cu}}^{+}}^{\text{o}}=0\text{.}15\text{V, }{\mathrm{E}}_{{\text{In}}^{2+}/{\text{In}}^{+}}^{\text{o}}=-0\text{.}40\text{V,}{ \mathrm{E}}_{{\mathrm{In}}^{3+}/{\mathrm{In}}^{+}}^{\mathrm{o}}=-0\text{.}42\text{V}$

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  $E_{F{e}^{3+}|F{e}^{2+}}^o=0.77V.)$

The Edison storage cell is represented as:

 $\mathrm{Fe}\left(\mathrm{s}\right)\left|\mathrm{FeO}\left(\mathrm{s}\right)\right|\mathrm{KOH}\left(\mathrm{aq}\right)|{\mathrm{Ni}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)|\mathrm{Ni}\left(\mathrm{s}\right)$

The half-cell reactions are:

 $\begin{array}{l}{\mathrm{Ni}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\underset{}{\rightleftharpoons }2\mathrm{NiO}\left(\mathrm{s}\right)+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=+0.40 \mathrm{V}\\ \mathrm{FeO}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\underset{}{\rightleftharpoons }\mathrm{Fe}\left(\mathrm{s}\right)+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=-0.87 \mathrm{V}\end{array}$

The cell e.m.f. and the maximum amount of electrical energy that can be obtained from one mole of ${\mathrm{Ni}}_2{\mathrm{O}}_3$?