Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

The cell potential for the following reaction is $0.3305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$, $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
The Gibbs free energy change for the above reaction at $298 \ K$ is $x \times 10^{-1} \ kJ \ mol^{-1}$;
The value of $x$ is ..... [nearest integer] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)