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09/01/2022 6:26 pm
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Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+(0.002 M) → Ni2+( (0.160 M) + 2Ag(s)
Given that E⊖cell = 1.05 V
1 Answer
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09/01/2022 6:31 pm
Applying Nernst equation we have,
Ecell = E⊖cell - \(\frac{0.0591}{n}\)log \(\frac{[Ni^{2+}]}{[Ag^+]^2}\)
= 1.05 - \(\frac{0.0591}{2}\)log \(\frac{0.160}{(0.002)^2}\)
= 1.05 - 0.02955 log \(\frac{0.16}{0.000004}\)
= 1.05 - 0.02955 log 4 × 104
= 1.05 - 0.02955 (log 10000 + log 4)
= 1.05 - 0.02955 (4 + 0.6021)
= 0.914 V