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Question
Calculate the emf of the cell in which the following reaction takes place:
\[\ce{Ni_{(s)} + 2Ag^+ (0.002 M) -> Ni^{2+} (0.160 M) + 2Ag_{(s)}}\]
Given that \[\ce{E^Θ_{cell}}\] = 1.05 V
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Solution
Applying the Nernst equation, we have:
`"E"_"cell" = "E"_"cell"^Θ - 0.0591/"n" log (["Ni"^(2+)])/(["Ag"^+]^2)`
= `1.05 "V" - 0.0591/2 log 0.160/(0.002)^2`
= `1.05 - 0.0591/2 log (4 xx 10^4)`
= `1.05 - 0.0591/2 (4.6021)`
= 1.05 V − 0.14 V
= 0.91 V
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