Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

For the disproportionation reaction $2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s ) + Cu ^{2+}( aq )$ at $298 \ K$,$\ln K$ (where $K$ is the equilibrium constant) is....... $\times 10^{-1}$.
Given: $(E _{ Cu ^{2+} / Cu ^{+}}^{0} = 0.16 \ V, E _{ Cu ^{+} / Cu }^{0} = 0.52 \ V, \frac{ RT }{ F } = 0.025 \ V)$

For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

For a reaction,$A_{(s)} + 2B_{(aq)}^{+} \rightleftharpoons A_{(aq)}^{2+} + 2B_{(s)}$,$K_{c}$ is $10^{12}$ at $25^{\circ} C$. The $E_{Cell}^{\circ}$ of the corresponding cell is $(F = 96500 \ C \ mol^{-1})$ (in $V$)

Calculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)

The reduction potential of a half cell consisting of a $Pt$ electrode immersed in $2.0 \ M \ Fe^{2+}$ and $0.02 \ M \ Fe^{3+}$ solution (in $V$) is. Given: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$