Electrode Potentials and Cell Potential

Two half-reactions of an electrochemical cell are given below :

 ${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right), \mathrm{E}°=+1.51\mathrm{V}$

 ${\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Sn}}^{4+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-},\mathrm{E}°=+0.15\mathrm{V}$

What is the standard cell potential?

Two half-cell reactions of an electrochemical cell are given below :

 ${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4\text{\hspace{0.17em}}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right),\mathrm{E}°=1.51\mathrm{V}$

 ${\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Sn}}^{4+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\mathrm{E}°=+0.15\mathrm{V}$

What would be the final cell potential?

A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential measured is 0.422 V. Determine the concentration of silver ion in the cell. Given :  ${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{o}}=+0.80\hspace{0.17em}\mathrm{V}, \hspace{0.17em}{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=+0.34\hspace{0.17em}\mathrm{V}$.

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

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

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

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

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  $\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}25°C.$

An acidic solution of ${\mathrm{Cu}}^{2+}$salt containing $0.4\mathrm{g}\hspace{0.17em}\mathrm{of}\hspace{0.17em}{\mathrm{Cu}}^{2+}$is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 mL and the current at 1.2 amp. The volume of gases evolved at NTP during the entire electrolysis:

A solution containing one mole per litre of each  $Cu{\left(N{O}_3\right)}_2;AgN{O}_3;H{g}_2{\left(N{O}_3\right)}_2;$  is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts are :

 $\begin{array}{l}\mathrm{Ag}/{\mathrm{Ag}}^{+}=+0.80,{\mathrm{Hg}}_2/{\mathrm{Hg}}_2^{2+}=+0.79\\ \mathrm{Cu}/{\mathrm{Cu}}^{2+}=+0.34,\mathrm{Mg}/{\mathrm{Mg}}^{2+}=-2.37\end{array}$

With increasing voltage, the sequence of deposition of metals on the cathode will be :

The equilibrium constant for the reaction,  $2{\mathrm{Fe}}^{3+}+3{\mathrm{I}}^{-}\underset{}{\rightleftharpoons }2{\mathrm{Fe}}^{2+}+{{\mathrm{I}}_3}^{-}$ would be, if the standard reduction potentials in acidic conditions are 0.77 V and 0.54 V respectively for  ${\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\mathrm{\hspace{0.17em}}{{\mathrm{I}}_3}^{-}/{\mathrm{I}}^{-}$couple.

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.