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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?
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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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