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Question
The standard potential of the cell in the following reaction is ______.
\[\ce{Cd_{(s)} + Cu^{2+}_{(1M)} -> Cd^{2+}_{(1M)} + Cu_{(s)}}\]
`("E"_("Cd")^circ = - 0.403V, "E"_("Cu")^circ = 0.334V)`
Options
– 0.737 V
0.737 V
– 0.069 V
0.069 V
Solution
The standard potential of the cell in the following reaction is 0.737 V.
Explanation:
`"E"_"cell"^circ = "E"_"RHE"^circ - "E"_"LHE"^circ`
= 0.334 – (– 0.403 V)
= 0.737 V
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