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Using the standard electrode potential, find out the pair between which redox reaction is not feasible. `E^(@)` values : `Fe^(3+)//Fe^(2+)=+0.77, I_(2)//I^(-)=+0.54 V`
`Cu^(2+)//Cu= + 0.34 V, Ag^(+)//Ag= +0.80 V`

A

`Fe^(3+)` and `I^(-)`

B

`Ag^(+)` and Cu

C

`Fe^(3+)` and Cu

D

Ag and `Fe^(3+)`

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The correct Answer is:
To determine which redox reaction is not feasible using the standard electrode potentials provided, we will analyze each pair of half-reactions based on their standard reduction potentials. ### Step-by-Step Solution: 1. **List the Standard Electrode Potentials**: - \( E^\circ \) for \( Fe^{3+} // Fe^{2+} = +0.77 \, V \) - \( E^\circ \) for \( I_2 // I^- = +0.54 \, V \) - \( E^\circ \) for \( Cu^{2+} // Cu = +0.34 \, V \) - \( E^\circ \) for \( Ag^+ // Ag = +0.80 \, V \) 2. **Rank the Electrode Potentials**: - The higher the standard reduction potential, the greater the tendency to gain electrons (be reduced). - Ranking from highest to lowest: - \( Ag^+ // Ag = +0.80 \, V \) (highest) - \( Fe^{3+} // Fe^{2+} = +0.77 \, V \) - \( I_2 // I^- = +0.54 \, V \) - \( Cu^{2+} // Cu = +0.34 \, V \) (lowest) 3. **Analyze Each Pair**: - **Pair 1: \( Fe^{3+} \) and \( I^- \)**: - For the reaction \( Fe^{3+} + I^- \rightarrow Fe^{2+} + I_2 \) to occur, \( Fe^{3+} \) must have a higher potential than \( I_2 \). - \( 0.77 > 0.54 \) (Reaction is feasible) - **Pair 2: \( Ag^+ \) and \( Cu \)**: - For the reaction \( Ag^+ + Cu \rightarrow Ag + Cu^{2+} \) to occur, \( Ag^+ \) must have a higher potential than \( Cu^{2+} \). - \( 0.80 > 0.34 \) (Reaction is feasible) - **Pair 3: \( Fe^{3+} \) and \( Cu \)**: - For the reaction \( Fe^{3+} + Cu \rightarrow Fe^{2+} + Cu^{2+} \) to occur, \( Fe^{3+} \) must have a higher potential than \( Cu^{2+} \). - \( 0.77 > 0.34 \) (Reaction is feasible) - **Pair 4: \( Ag \) and \( Fe^{3+} \)**: - For the reaction \( Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+} \) to occur, \( Fe^{3+} \) must have a higher potential than \( Ag^+ \). - \( 0.77 < 0.80 \) (Reaction is NOT feasible) 4. **Conclusion**: - The pair between which the redox reaction is not feasible is \( Ag \) and \( Fe^{3+} \). ### Final Answer: The redox reaction between \( Ag \) and \( Fe^{3+} \) is not feasible.
To determine which redox reaction is not feasible using the standard electrode potentials provided, we will analyze each pair of half-reactions based on their standard reduction potentials. ### Step-by-Step Solution: 1. **List the Standard Electrode Potentials**: - \( E^\circ \) for \( Fe^{3+} // Fe^{2+} = +0.77 \, V \) - \( E^\circ \) for \( I_2 // I^- = +0.54 \, V \) - \( E^\circ \) for \( Cu^{2+} // Cu = +0.34 \, V \) ...
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Using the standard electrode potential, find out the pair between which redox reaction is not feasible. E^(θ) values: Fe^(3+)//Fe^(2+)=0.77, I_(2)//I^(-)=+0.54 , Cu^(2+)//Cu=+0.34, Ag^(+)//Ag=+0.80V

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The standard reduction potential data at 25^(@)C is given below E^(@) (Fe^(3+), Fe^(2+)) = +0.77V , E^(@) (Fe^(2+), Fe) = -0.44V , E^(@) (Cu^(2+),Cu) = +0.34V , E^(@)(Cu^(+),Cu) = +0.52 V , E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V , E^(@) (Cr^(3+), Cr) =- 0.74V , E^(@) (Cr^(2+),Cr) = - 0.91V , Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: {:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):} Codes: