For the reaction $A_{(s)} + 2B^{+}_{(aq)} \rightarrow A^{2+}_{(aq)} + 2B_{(s)}$,the value of $K_c$ is $10^{12}$. What is the value of $E^o_{cell}$ (in $, V$)?

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

The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:
$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

The $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)

At $298 \ K$,some standard electrode potentials are given below:

$Pb^{2+} / Pb$	$-0.13 \ V$
$Ni^{2+} / Ni$	$-0.24 \ V$
$Cd^{2+} / Cd$	$-0.40 \ V$
$Fe^{2+} / Fe$	$-0.44 \ V$

Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$

At $298 \, K$,the standard reduction potential for $Cu^{2+}/Cu$ electrode is $0.34 \, V$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.