JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the cell \(Pt _{( s )}\left| H _2( g , 1\,atm )\right| H ^{+}( aq , 1 M )|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid Pt ( s )\) When the potential of the cell is \(0.712\,V\) at \(298\,K\), the ratio \(\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right]\) is \(.......\)(Nearest integer) Given: \(Fe ^{3+}+ e ^{-}= Fe ^{2+}, E ^{\circ} Fe ^{3+}, Fe ^{2+} \mid Pt =0.771\) \(\frac{2.303 RT }{ F }=0.06\,V\)
- A \(30\)
- B \(10\)
- C \(20\)
- D \(50\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(Pt _{( s )}\left| H _2( g , 1 atm )\right| H ^{+}( aq , 1 M ) \| Fe ^{3+}( aq ), Fe ^{2+}( aq )| Pt |( s )\) \(\text { at anode } H _2 \longrightarrow 2 H ^{+}+2 e ^{-}\) \(\text {At cathode } Fe _{ aq }^{3+}+ e ^{-} \longrightarrow Fe _{ aq }^{2+}\)…
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