Answer:
The reaction of I- with Fe3+ in acidic medium
The iodide is oxidised to iodine and Fe3+ is reduced to Fe2+
Thus the reduction half reaction is
Fe3+ + 1 e- Fe2+ ------------------ (1)
The oxidation half reaction is
2 I- I2 + 2 e- --------------- (2)
The overall reaction is obtained by multiplying equation (1) by 2 and add to equation (2)
[ This is to balance the number of electrons on both sides.]
2 Fe3+ + 2 I- I2 + 2 Fe2+ ------------------ (3)
This equation (3) is the balanced overall redox reaction
It can be expressed in cell notation as
/oxidation//reduction/ or /anode cell//cathode cell/
that is 2I- / I2 // Fe3+ / Fe2+
anode cathode
(oxidation) ( reduction)
E0cell = Eox (anode) + Ered (cathode)
From the standard reduction potential table
the reduction potential for I2 + 2 e- 2I- E = 0.54 V
Therefore the oxidation potential = -0.54 V
the reduction potential for Fe3+ + e- Fe2+ E = 0.77 V
Therefore E0cell = -0.54 + 0.77 = + 0.23 V
Since E0cell is positive , G0 = -nFE0 will be negative and hence the forward reaction is spontaneous.
E0cell for the reverse reaction is -0.23 V
Since E0cell is negative , G0 = -nFE0 will be positive and hence the reverse reaction is non-spontaneous.
Thus the reaction
2 Fe3+ + 2 I- I2 + 2 Fe2+ (feasible)
I2 + 2 Fe2+ 2 Fe3+ + 2 I- ( not feasible)
is occurence in the forward direction. The reverse reaction is non-occurence.
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