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 $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

Find the $emf$ of the following cell reaction,given $E^0_{Cr^{3+}/Cr} = -0.74 \ V$ and $E^0_{Fe^{2+}/Fe} = -0.44 \ V$ at $25^{\circ} C$ for the cell: $Cr \ | \ Cr^{3+}(0.1 \ M) \ || \ Fe^{2+}(0.01 \ M) \ | \ Fe$. (in $V$)

The standard Gibbs energy change in $kJ \ mol^{-1}$ for a galvanic cell $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$ that has a standard emf of $0.5 \ V$ is: $\left(F = 96500 \ C \ mol^{-1}\right)$

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
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] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

Find the observed $EMF$ of the cell $Cd | Cd^{2+}(0.01 \ M) || Cu^{2+}(0.01 \ M) | Cu$ under conditions where the internal resistance is $4 \ \Omega$ and it is producing a current of $0.15 \ A$. (Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.35 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.4 \ V$) (in $V$)