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प्रश्न
Calculate the emf of the following cell:
\[\ce{Ni_{(s)} + 2Ag+(0.01 M) -> Ni^{2+} (0.1 M) + 2Ag_{(s)}}\]
Given that `E_(cell)^0` = 1.05 V, log 10 = 1
संख्यात्मक
उत्तर
Using the Nernst equation:
`E_(cell) = E_(cell)^0 - 0.0592/n log [["Products"]]/[["Reactants"]]`
= `1.05 - 0.0592/2 log [["Ni"^(2+)]]/["Ag"^+]^2`
= `1.05 - 0.0291 log 0.1/(0.01)^2`
= 1.05 − 0.0291 log 103 ...[∴ log 10 = 1]
= 1.05 − 0.0291 × 3
= 1.05 − 0.0873
Ecell = 0.962 V
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