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Answer the following in one or two sentences. What is standard cell potential for the reaction 3NiA(s)+2AlA3+ (1M)⟶3NiA2+ (1M)+2Al(s), if NiENi∘ = –0.25 V and EAlEAl∘ = –1.66 V?

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Question

Answer the following in one or two sentences.

What is standard cell potential for the reaction

\[\ce{3Ni_{(s)} + 2Al^{3+} (1M) → 3Ni^{2+} (1M) + 2Al(s)}\], if `E_"Ni"^circ` = –0.25 V and  `"E"_("Al")^circ` = –1.66 V?

Short/Brief Note
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Solution

Given: `"E"_("Ni")^circ` = –0.25 V, `"E"_("Al")^circ` = – 1.66 V

To find: Standard cell potential

Formula: `E_"cell"^circ = E_"cathode"^circ - E_"anode"^circ`

Calculation: Electrode reactions are

At anode: \[\ce{Ni_{(s)} -> Ni^{2+}_{ (aq)} + 2e^-}\]

At cathode: \[\ce{Al^{3+}_{ (aq)} + 3e^{-} -> Al_{(s)}}\]

The standard electrode potential is given by

`"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`

`"E"_"cell"^circ = "E"_"Al"^circ - "E"_"Ni"^circ`

= (–1.66 V) – (–0.25 V)

= –1.41 V

The standard cell potential for the reaction is –1.41 V.

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