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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

\(Pt ( s ) H _2( g )(1 bar )\left| H ^{+}( aq )(1 M )\right|\left| M ^{3+}( aq ), M ^{+}( aq )\right| Pt ( s )\) The \(E _{\text {ceil }}\) for the given cell is \(0.1115\,V\) at \(298\,K\) when \(\frac{\left[ M ^{+}( aq )\right]}{\left[ M ^{3+}( aq )\right]}=10^{ a }\) The value of a is Given : \(E _{ M ^{3+} / M ^{+}}=0.2\,V\) \(\frac{2.303 RT }{ F }=0.059\,V\)

  1. A \(2\)
  2. B \(6\)
  3. C \(8\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

Overall reaction :- \(H _{2( g )}+ M _{( aq )}^{3+} \longrightarrow M _{( aq )}^{+}+2 H _{( aq )}^{+}\) \(E _{\text {Coll }}= E _{\text {Cathode }}^{\circ}- E _{\text {amode }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ M ^{+}\right] \times 1^2}{\left[ M ^{+3}\right] 1}\)…
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