Question

Using the following reduction potentials:

I₂(s) + 2e^- <---> 2I^-(aq) E⁰ = 0.535 V

I₂ (aq) +2e^- <---> 2I^-(aq) E⁰ = 0.620 V

I₃^-(aq) + 2e^- <---> 3I^-(aq) E⁰ = 0.535 V

a) Calculate the equilibrium constant for I₂(aq) + I^-(aq) <---> I₃^-(aq)

b) Calculate the equilibrium constant for I₂(s) + I^-(aq) <---> I₃^-(aq)

c) Calculate the solubility (g/L) of I₂(s) in water.

Answer 1

Standard potential (E⁰) and equilibrium constant (K) are related by relation,

E⁰ = (0.059/n) ln(K) ……………….( After substituting R, T, F values)

Let K1, K2, K3 be the equilibrium constants for following 3 transformations respectively

For these reductions Standard EMF (E⁰) are given, using which let us find out respective equilibrium constant for each of them,

a) I₂(s) + 2e- <---> 2I-(aq.)

K1 = [I^-]² ……………….(1) ………..(I2 (s) do not appear in rate expression)

[I^-] = (K1)^1/2 ……….(Taking sq. root of both sides) …………….(1a)

E0 = 0.535 V, n = 2,

0.535 = (0.059 / 2) ln(K1)

0.535 = 0.0295 ln(K1)

ln(K1) = 0.535 / 0.0295

ln(K1) = 18.14

K1 = e^18.14

K1 = 7.55 x 10⁷ ………. (2)

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b) I₂ (aq) +2e- <---> 2I-(aq)

K2 = [I^-]² / [I₂] ……….. (3)

K2 = K1 / [I₂] ……..(by eq. 1) …………..(3a)

E⁰ = 0.620 V, n = 2

0.620 = 0.0295 ln(K2)

ln(K2) = 0.620 / 0.0295

ln(K2) = 21.02

K2 = e^21.02

K2 = 1.35 x 10⁹ ……….. (4)

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c) I₃^-(aq) + 2e- <---> 3I-(aq)

K3 = [I^-]³ / [I₃^-] …… (5)

K3 = (K1)^3/2 / [I₃^-] ……………...(By eq. 1a) …………..(5a)

E⁰ = 0.535 V , n = 2,

0.535 = 0.0295 ln(K3)

ln(K3) = 0.535 / 0.0295

ln(K3) = 18.14

K3 = 7.55 x 10⁷……………. (6)

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a)Let K4 be the equilibrium for the transformation,

I₂(aq) + I^-(aq) <---> I₃^-(aq)

K4 = [I₃^-] / [I₂][I-]

[I₃^-] = (K1)^3/2/ (K3) ………… (by rearranging eq.5a)

[I₂] = K1/K2 ………….(by rearranging eq.3a)

[I^-] = (K1)^1/2…….(by eq. 1a)

Put all these value in expression for K4,

K4 = [(K1)^3/2/ (K3)] / [K1/K2][ (K1)^1/2]

K4 = [(K1)^3/2/ (K3)] / [(K1)^3/2/K2]

K4 = (1/K3) / (1/K2) ………………[(K2)3/2 cancelled out]

K4 = K2 / K3

K4 = 1.35 x 10⁹/ 7.55 x 10⁷

K4 = 17.88

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b) Calculate the equilibrium constant for I₂(s) + I^-(aq) <---> I₃^-(aq)

Let K5 be the equilibrium constant for this transformation,

K5 =[I₃^-] / [I^-]

[I₃^-] = (K1)^3/2/ (K3) ………… (by rearranging eq.5a)

[I^-] = (K1)^1/2…….(by eq. 1a)

Let us put these value in expression for K5,

K5 = [(K1)^3/2/ (K3) ] / (K1)^1/2

K5 = (K1)² / K3

K5 = (7.55 x 10⁷ )² /(7.55 x 10⁷)

K5 = 7.55 x 10⁷.

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c) Calculate the solubility (g/L) of I2(s) in water

Let ‘S’ be the solubility of I₂ in mole/L in water,

I₂(s) -------- > 2I^- (aq.)

Solubility of I2 is ‘S’ mole/L

Hence, [I^-] = (2S) mole/L

Ksp = [I-]² = (2S)­²

Ksp = K1 = 4S²

7.55 x 10⁷ = 4S²

S2 = 7.55 x 10⁷ /4

S2 = 1.8875 x 10⁷

Taking square roots of both sides,

S = 4.345 x 10³ mole/L

To above value if we multiply by molar mass of I₂ we will have solubility in g/L

Molar mass of I2 = 253.809 g

Hence,

S = 4.345 x 10³ x 253.809 g/L

S = 1.103 x 10⁶ g/L

Solubility of I2(s) in water is 1.103 x 10⁶ g/L.

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Note : I used many times laws of exponent here. If you find it difficult to understand please let me know.