ExamBro
ExamBro
JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

The electrode potential of the following half cell at \(298\,K\) \(X \left| X ^{2+}(0.001 M ) \| Y ^{2+}(0.01 M )\right| Y\) is \(.......\times 10^{-2} V\) (Nearest integer). Given: \(E _{ x ^{2+} \mid x }^0=-2.36\,V\) \(E _{ Y ^{3+} \mid Y }^0=+0.36\,V\) \(\frac{2.303\,RT }{ F }=0.06\,V\)

  1. A \(274\)
  2. B \(273\)
  3. C \(272\)
  4. D \(275\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(275\)

Step-by-step Solution

Detailed explanation

\(X + Y ^{2+} \rightarrow Y + X ^{2+}\) \(E _{\text {Cell }}^0=0.36-(-2.36)=2.72\,V\) \(E _{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{0.001}{0.01}\) \(=2.72+0.03=2.75\,V\) \(=275 \times 10^{-2}\,V\)