NEET · Chemistry · STD 12 - 2. Electrochemistry
Find the emf of the cell in which the following reaction takes place at \(298 \,K\) (In \(V\)) \(Ni ( s )+2 Ag ^{+}(0.001 M ) \rightarrow Ni ^{2+}(0.001 M )+2 Ag ( s )\) \((\) Given that \(E _{\text {cell }}^{\circ}=10.5 \,V , \frac{2.303 RT }{ F }=0.059\) at \(298\, K )\)
- A \(1.385\)
- B \(10.4115\)
- C \(1.05\)
- D \(1.0385\)
Answer & Solution
Correct Answer
(B) \(10.4115\)
Step-by-step Solution
Detailed explanation
\(Ni ( s )+2 Ag ^{+}(0.001 \,M ) \rightarrow Ni ^{+2}(0.001\, M )+2 Ag ( s )\) \(E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.059}{ n } \log \frac{\left[ Ni ^{+2}\right]^{1}}{\left[ Ag ^{+}\right]^{2}}\)
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