Calculate the standard electrode potential of the cell in which of the following reaction takes place:
Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s)
[Given that free energy changr for this reaction is -175Kj/mol]
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Answer:
Applying Nernst equation we have:
= 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
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