The reduction potential of a hydrogen electrode at $pH = 1$ is ........ $V$.

Determine the electrode potential of $Sn^{2+}(0.01 \text{ M}) | Sn(s)$ at $25^{\circ} \text{C}$ if $E^0_{Sn^{2+}/Sn}$ is $-0.136 \text{ V}$. (in $\text{ V}$)

In a cell reaction,a $2$ electron change occurs. The standard $e.m.f.$ of the cell at $25^\circ C$ is $0.295 \ V$. What is the equilibrium constant for the reaction at $25^\circ C$?

Determine the electrode potential of $Sn^{2+} (0.01 \text{ M}) | Sn(s)$ at $25^\circ \text{C}$ if $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$. (in $\text{ V}$)

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

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 $V$)