The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

Find the $emf$ of the following cell reaction,given $E^0_{Cr^{3+}/Cr} = -0.74 \ V$ and $E^0_{Fe^{2+}/Fe} = -0.44 \ V$ at $25^{\circ} C$ for the cell: $Cr \ | \ Cr^{3+}(0.1 \ M) \ || \ Fe^{2+}(0.01 \ M) \ | \ Fe$. (in $V$)

Determine the electrode potential of $Sn^{2+}(0.01 \text{ M}) | Sn(s)$ at $25^{\circ} \text{C}$ if $E^0_{Sn^{2+}/Sn}$ is $-0.136 \text{ V}$. (in $\text{ V}$)

Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

The following reaction takes place at $298 \, K$ in an electrochemical cell involving two metals $A$ and $B$,
$A^{2+}_{(aq)} + B_{(s)} \rightarrow B^{2+}_{(aq)} + A_{(s)}$
with $[A^{2+}] = 4 \times 10^{-3} \, M$ and $[B^{2+}] = 2 \times 10^{-3} \, M$ in the respective half-cells,the cell $EMF$ is $1.091 \, V$.
The equilibrium constant of the reaction is closest to

At $298 \, K$,find out the $emf$ for the cell: $Al_{(s)} | Al^{+3} (0.1 \, M) || Fe^{+2} (0.001 \, M) | Fe_{(s)}$. Given $E^o_{Al^{+3}/Al} = -1.66 \, V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \, V$. (in $, V$)