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: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

When is the reduction potential of a hydrogen half-cell negative?

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

If a cell has a standard electrode potential of $0.295 \ V$ and $n = 2$,calculate its equilibrium constant at $298 \ K$.

For the electrochemical cell shown below:
$Pt \mid H_{2}(p=1 \, atm) \mid H^{+}(aq., x \, M) \mid\mid Cu^{2+}(aq., 1.0 \, M) \mid Cu_{(s)}$
The potential is $0.49 \, V$ at $298 \, K$. The $pH$ of the solution is closest to:
[Given: Standard reduction potential,$E^{\circ}$ for $Cu^{2+}/Cu$ is $0.34 \, V$; Gas constant,$R = 8.31 \, J \, K^{-1} \, mol^{-1}$; Faraday constant,$F = 9.65 \times 10^{4} \, J \, V^{-1} \, mol^{-1}$]