Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)

For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

The standard e.m.f. of a cell,involving one electron change,is found to be $0.591 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction is: $(F = 96,500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

By how much would the oxidising power of the $(MnO_4^- / Mn^{2+})$ couple change if the $H^{+}$ ion concentration is increased up to $100$ times at $25 ^oC$?

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 $Fe$