Problem: Consider a galvanic cell that uses the reaction4H3O+ (aq) + 4I- (aq) + O2 (g) → 2I2 (s) 6H2O (l).If Ecell is +0.74 V, calculate ΔG°f of I- (aq) at 25 °C. Show all work and circle your final answer. ΔG°f (H2O(l)) = -237.18 kJ/mol at 25 °C.Anode: 4I-(aq) → 2I2(g) + 4e-                                                    ε°ox = =0.54 VCathode: O2(g) + 4H3O+(aq) + 4e- → 6H2O(l)                         ε°red = +1.23 VOverall rx: 4H3O+(aq) + 4I-(aq) + O2(g) → 2I2(s) + 6H2O(l)      ε°cell = 0.69 V

FREE Expert Solution

For this problem, we have to calculate for the ΔGf of I- when Ecell for the reaction is +0.74 V

Recall that ΔG can be calculated as:

°G = °Gf products - °Gf reactants


In this case, we can calculate for the ΔGf of I- using the equation as:


°G =[(2 mol)°Gf (I2)+(6 mol)°Gf (H2O)]                                  - [(4 mol)°Gf (H3O+)+(4 mol)°Gf (I-)+(1 mol)°Gf (O2)]°G =[(6 mol)°Gf (H2O)]- [(4 mol)°Gf (I-)]


Note that ΔGf for compounds in standard conditions are 0 (I2, H3O+ and O2)

To do so, we have to follow the steps:

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Problem Details

Consider a galvanic cell that uses the reaction

4H3O+ (aq) + 4I- (aq) + O2 (g) → 2I2 (s) 6H2O (l).

If Ecell is +0.74 V, calculate ΔG°f of I- (aq) at 25 °C. Show all work and circle your final answer.

 ΔG°(H2O(l)) = -237.18 kJ/mol at 25 °C.

Anode: 4I-(aq) → 2I2(g) + 4e-                                                    ε°ox = =0.54 V

Cathode: O2(g) + 4H3O+(aq) + 4e- → 6H2O(l)                         ε°red = +1.23 V

Overall rx: 4H3O+(aq) + 4I-(aq) + O2(g) → 2I2(s) + 6H2O(l)      ε°cell = 0.69 V

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