Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible. The \( E^\circ \) values are: \[ {Fe}^{3+}/{Fe}^{2+} = +0.77, \quad {I}_2/{I}^- = +0.54 \] \[ {Cu}^{2+}/{Cu} = +0.34, \quad {Ag}^+/{Ag} = +0.80 \, {V} \]

• A. \( {Fe}^{3+} \) and \( {I}^- \)
• B. \( {Ag}^+ \) and \( {Cu} \)
• C. \( {Fe}^{3+} \) and \( {Cu} \)
• D. \( {Ag} \) and \( {Fe}^{3+} \)

Hint:

For a redox reaction to be feasible, the standard electrode potential \( E^\circ \) must be positive. If it's negative, the reaction is non-spontaneous.

Answer 1

The Correct Option is D

For a redox reaction to be feasible, the standard electrode potential for the overall reaction should be positive. This can be determined by subtracting the standard electrode potential of the reduction half-reaction from the oxidation half-reaction. If the result is positive, the reaction is spontaneous. Let's check each pair: - For \( {Fe}^{3+} \) and \( {I}^- \): \[ E^\circ = E^\circ_{{reduction}} ({Fe}^{3+}/{Fe}^{2+}) - E^\circ_{{oxidation}} ({I}^-/{I}_2) \] \[ E^\circ = 0.77 - (-0.54) = +1.31 \, {V} \quad ({Reaction is feasible}) \] - For \( {Ag}^+ \) and \( {Cu} \): \[ E^\circ = E^\circ_{{reduction}} ({Ag}^+/ {Ag}) - E^\circ_{{oxidation}} ({Cu}^{2+}/{Cu}) \] \[ E^\circ = 0.80 - 0.34 = +0.46 \, {V} \quad ({Reaction is feasible}) \] - For \( {Fe}^{3+} \) and \( {Cu} \): \[ E^\circ = E^\circ_{{reduction}} ({Fe}^{3+}/ {Fe}^{2+}) - E^\circ_{{oxidation}} ({Cu}^{2+}/ {Cu}) \] \[ E^\circ = 0.77 - 0.34 = +0.43 \, {V} \quad ({Reaction is feasible}) \] - For \( {Ag} \) and \( {Fe}^{3+} \): \[ E^\circ = E^\circ_{{reduction}} ({Fe}^{3+}/ {Fe}^{2+}) - E^\circ_{{oxidation}} ({Ag}/{Ag}^+) \] \[ E^\circ = 0.77 - 0.80 = -0.03 \, {V} \quad ({Reaction is not feasible}) \] Thus, the redox reaction between \( {Ag} \) and \( {Fe}^{3+} \) is not feasible because the \( E^\circ \) value is negative.