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Question
Answer the following in one or two sentences.
What is the standard cell potential for the reaction?
\[\ce{2Al(s) + 3Ni^{2⊕}(1M) -> 2Al^{3⊕}(1 M) + 3Ni(s)}\]
if `"E"_"Ni"^0` = −0.25 V and `"E"_"Al"^0` = −1.66 V
Solution
Given: `"E"_"Ni"^0` = −0.25 V, `"E"_"Al"^0` = −1.66 V
To find: Standard cell potential
Formula: `"E"_"cell"^0 = "E"_"cathode"^0 - "E"_"anode"^0`
Calculation: Electrode reactions are
At anode: \[\ce{Al_{(s)} -> Al^{3+}_{( aq)} + 3e^-}\]
At cathode: \[\ce{Ni^{2+}_{ (aq)} + 2e^- -> Ni_{(s)}}\]
The standard electrode potential is given by
`"E"_"cell"^0 = "E"_"cathode"^0 - "E"_"anode"^0`
`"E"_"cell"^0 = "E"_"Ni"^0 - "E"_"Al"^0`
= (–0.25 V) – (−1.66 V)
= 1.41 V
The standard cell potential for the reaction is 1.41 V.
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