JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
In a cell, the following reactions take place \(Fe ^{2+} \rightarrow Fe ^{3+} + e ^{-} \quad E _{ Fe ^{3+} / Fe ^{2+}}=0.77 \,V\) \(2 I ^{-} \rightarrow I _{2}+2 e ^{-} \quad E _{ I _{2} / I ^{-}}^{0}=0.54 \,V\) The standard electrode potential for the spontaneous reaction in the cell is \(x \times 10^{-2} \,V 298\) \(K\). The value of \(x\) is .... (Nearest Integer)
- A \(95\)
- B \(202\)
- C \(23\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(23\)
Step-by-step Solution
Detailed explanation
\(Fe _{\text {Cathode }}^{+3}+ I _{\text {anode }}^{-} \longrightarrow I _{2}+ Fe ^{+2}\) \(E _{\text {Cell }}^{0}= E _{\text {cathode }}^{0}- E _{\text {anode }}^{0}\) \(=0.77-0.54\) \(=0.23\) \(=23 \times 10^{-2}\, V\)
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