JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the cell \(Pt ( s )\left| H _2( s )( latm )\right| H ^{+}\left( aq ,\left[ H ^{+}\right]=1\right)|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid \operatorname{Pt}( s )\) Given : \(E _{ Fe ^{3+} / e ^{2 *}}^0=0.771\,V\) and \(E _{ H ^{+}+\frac{1}{2} H _2}^0=0 V , T =298\,K\) If the potential of the cell is \(0.712\,V\) the ratio of concentration of \(Fe ^{2+}\) to \(Fe ^{2+}\) is \(........\).(Nearest integer)
- A \(100\)
- B \(10\)
- C \(105\)
- D \(852\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} H _2( g )+ Fe ^{3+} \text { (aq.) } \longrightarrow H ^{+}( aq )+ Fe ^{2+}( aq .)\) \(E = E ^{\circ}-\frac{0.059}{1} \log \frac{\left[ Fe ^{2+}\right]}{\left\lfloor Fe ^{3+}\right\rfloor}\)…
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