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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 \[\ce{E{^{\circ}_{Ni}}}\] = −0.25 V and \[\ce{E{^{\circ}_{Al}}}\] = −1.66 V?
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Solution
Given: \[\ce{E{^{\circ}_{Ni}}}\] = −0.25 V, \[\ce{E{^{\circ}_{Al}}}\] = −1.66 V
To find: Standard cell potential
Formula: \[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
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
\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{Ni}} - E{^{\circ}_{Al}}}\]
= (–0.25 V) – (−1.66 V)
= 1.41 V
∴ The standard cell potential for the reaction is 1.41 V.
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