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:

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

Calculate the concentration of $Cu^{2+}$ in a $Cu$ plate kept in a $0.2 \ M$ $CuSO_4$ solution when the potential becomes $0.0 \ V$.

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)