Calculate the standard cell potential for a cell having the following reaction: $2 Al_{(s)} + 3 Ni^{2+} \rightarrow 2 Al^{3+} + 3 Ni_{(s)}$ given that $E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$ and $E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$. (in $V$)

If the standard potential of the electrochemical cell for the reaction $2Ag_{(aq)}^+ + Cd_{(s)} \to Cd_{(aq)}^{2+} + 2Ag_{(s)}$ is $1.20 \ V$,calculate the standard Gibbs free energy change in $kJ$.

$MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298 \ K$: $MX_{(s)} \rightleftharpoons M^{+}_{(aq)} + X^{-}_{(aq)}$; $K_{sp} = 10^{-10}$. If the standard reduction potential for $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ is $(E^{\ominus}_{M^{+}/M}) = 0.79 \ V$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\ominus}_{X^{-}/MX_{(s)}/M}$ is . . . . . . $mV$. (nearest integer) [Given: $\frac{2.303 RT}{F} = 0.059 \ V$]

Given below are the half-cell reactions: $Mn^{2+} + 2e^- \rightarrow Mn$ $(E^0 = -1.18 \text{ V})$; $Mn^{3+} + e^- \rightarrow Mn^{2+}$ $(E^0 = +1.51 \text{ V})$. The $E^0_{cell}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be . . . . . .

Given $:$
$(i) \, Cu^{2+} + 2e^- \rightarrow Cu \,, \, E^o = 0.337 \, V$
$(ii) \, Cu^{2+} + e^- \rightarrow Cu^{+} \,, \, E^o = 0.153 \, V$
Electrode potential,$E^o$ for the reaction,
$Cu^{+} + e^- \rightarrow Cu \,,$ will be $............$ $V$.

Standard electrode potential of $NHE$ at $298\,K$ is .............. $V$.