If standard reduction potential $\left(\mathrm{E}^{\circ}\right)$ of $\left(\stackrel{+2}{\mathrm{Ni}_{(\mathrm{aq})}} \mid \mathrm{Ni}_{(\mathrm{s})}\right)$ and $\left(\underset{(\mathrm{aq})}{\stackrel{+3}{\mathrm{Al}}} \mid \mathrm{Al}_{(\mathrm{s})}\right)$ are -0.25 V and -1.66 V respectively. What is standard emf of cell reaction $2 \mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Ni}_{(\mathrm{aq})}^{+2} \rightarrow 2 \mathrm{Al}_{(\mathrm{aq})}^{+3}+3 \mathrm{Ni}_{(\mathrm{s})}$

Standard EMF of the cell reaction

The cell reaction is $2\textrm{Al}_{(\textrm{s})} + 3\textrm{Ni}_{(\textrm{aq})}^{+2} \rightarrow 2\textrm{Al}_{(\textrm{aq})}^{+3} + 3\textrm{Ni}_{(\textrm{s})}$.

Aluminum is oxidized at the anode, $\textrm{Al} \rightarrow \textrm{Al}^{+3} + 3\textrm{e}^{-}$, and nickel is reduced at the cathode, $\textrm{Ni}^{+2} + 2\textrm{e}^{-} \rightarrow \textrm{Ni}$.

Given standard reduction potentials: $\textrm{E}^{\circ}(\textrm{Ni}^{+2} | \textrm{Ni}) = -0.25\textrm{ V}$ and $\textrm{E}^{\circ}(\textrm{Al}^{+3} | \textrm{Al}) = -1.66\textrm{ V}$.

The cell EMF is calculated as $\textrm{E}^{\circ}_{\textrm{cell}} = \textrm{E}^{\circ}_{\textrm{cathode}} - \textrm{E}^{\circ}_{\textrm{anode}} = (-0.25) - (-1.66) = +1.41\textrm{ V}$.

$\textrm{E}^{\circ}_{\textrm{cell}} = +1.41\textrm{ V}$.