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

$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)

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: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$

What is the value of $E_{cell}$ at $298 \ K$ for the reaction,$Zn_{(s)} + Cu^{+2}(0.1 \ M) \rightarrow Zn^{+2}(0.1 \ M) + Cu_{(s)}$ if $E^{\circ}_{cell} = 1.1 \ V$ (in $V$)?

Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$

The standard $EMF$ for the given cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$ is $1.10 \ V$ at $25^oC$. The $EMF$ for the cell reaction,when $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,at $25^oC$ is .......... $V$.