Chemistry - Electrochemistry Question with Solution | TestHub

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

In this reaction, oxidation of $\mathrm{Al}$ takes place, so it acts as the anode. Reduction of ${\mathrm{Ni}}^{+2}$ takes place, so it acts as the cathode.

The standard emf of the cell is calculated using the formula:

$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$

Substituting the given standard reduction potentials:

$E_{\text{cell}}^{\circ }=E_{{\mathrm{Ni}}^{+2}/\mathrm{Ni}}^{\circ }-E_{{\mathrm{Al}}^{+3}/\mathrm{Al}}^{\circ }$

$E_{\text{cell}}^{\circ }=-0.25-(-1.66)$

$E_{\text{cell}}^{\circ }=-0.25+1.66=+1.41\text{ V}$

Answer: $+1.41\mathrm{V}$