Dear Student, The correct answer is (1)2I- → I2 + 2e- E°oxidation =-0 536 VFe3+ + e- → Fe2+ E°reduction =-0 77 V2Fe3++2I- → 2Fe2+ + I2E°= E°oxidation + E°reduction = -0 536-(-0 77) =0 234 V which is+veSo, reaction takes place

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Dear Student,

The correct answer is (1) 2 I^{-} \to I_{2} + 2 e^{-} E °_{oxidation} = - 0.536 V {Fe}^{3 +} + e^{-} \to {Fe}^{2 +} E °_{reduction} = - 0.77 V 2 {Fe}^{3 +} + 2 I^{-} \to 2 {Fe}^{2 +} + I_{2} E ° = E °_{oxidation} + E °_{reduction} = - 0.536 - (- 0.77) = 0.234 V which is + ve So, reaction takes place

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