At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

What is the standard free energy change for the cell,having the following cell reaction (in $kJ$)?
$2 Ag_{(aq)}^{+} + Cd_{(s)} \longrightarrow 2 Ag_{(s)} + Cd_{(aq)}^{2+}, E^{\circ}_{cell} = 1.20 \ V$

For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

In the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)

For a given half-cell,$Al^{3+} + 3e^{-} \rightarrow Al$,on increasing the concentration of aluminium ions,the electrode potential will