For the cell involving the following reaction: $Zn_{(s)} + Ni^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$. Given $E^{\circ}_{\text{cell}} = 0.5 \ V$. What is the standard Gibbs energy change of the cell reaction (in $kJ$)?

Using the standard electrode potentials given below,identify the correct statements from the following.
$Fe^{2+} + 2e^{-} \longrightarrow Fe ; E^{\circ} = -0.44 \ V$
$Cu^{2+} + 2e^{-} \longrightarrow Cu ; E^{\circ} = +0.34 \ V$
$Ag^{+} + e^{-} \longrightarrow Ag ; E^{\circ} = +0.80 \ V$
$(i)$ Copper can displace iron from $FeSO_4$ solution.
$(ii)$ Iron can displace copper from $CuSO_4$ solution.
$(iii)$ Silver can displace copper from $CuSO_4$ solution.
$(iv)$ Iron can displace silver from $AgNO_3$ solution.

If the standard reduction potentials of $Zn$, $Ni$, and $Fe$ are $-0.76 \text{ V}$, $-0.23 \text{ V}$, and $-0.44 \text{ V}$ respectively, determine the electrodes $X$ and $Y$ for the reaction $X(s) + Y^{+2}_{(aq)} \rightarrow X^{+2}_{(aq)} + Y(s)$ to be spontaneous.

Given: $E^{0}_{Mn^{+7} \mid Mn^{+2}} = 1.5 \ V$ and $E^{0}_{Mn^{+4} \mid Mn^{+2}} = 1.2 \ V$,then $E^{0}_{Mn^{+7} \mid Mn^{+4}}$ is (in $V$)

On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

Calculate the $E_{cell}^o$ for $Zn_{(s)}|Zn_{(1M)}^{2+}| |Cd_{(1M)}^{2+}|Cd_{(s)}$ at $25^{\circ} C$ given that $E_{Zn^{2+}/Zn}^{\circ} = -0.763 \ V$ and $E_{Cd^{2+}/Cd}^{\circ} = -0.403 \ V$. (in $V$)