Question

Consider the reaction Cr₂O₇²⁻(aq) + I⁻(aq) → Cr³⁺(aq) + I₂(s).

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Answer 1

I₂/I^- is anode , E⁰ I₂/I^-  = 0.54 V

Anode reaction;

2I^- - 2e $\small \to$ I₂ (1)

and Cr₂O₇^2-/Cr³⁺ is cathode, E⁰Cr₂O₇^2-/Cr³⁺ = 1.33 V

Cathode reaction;

Cr₂O₇^2- + 14H⁺ + 6 e $\small \to$ 2 Cr³⁺ + 7H₂O (2)

Eq. 1 * 3 + Eq. 2

Cr₂O₇^2- + 14H⁺ + 6 I^-  $\small \to$ 2 Cr³⁺ + 7H₂O + 3 I₂

number of electron transferred (n) = 6

E⁰_cell = E^o_cathode - E⁰_anode = 1.33 - 0.54 = 0.79 V

$\small \Delta G$⁰ = - nFE⁰_cell

= - 6 $\small \times$ 96500 $\small \times$ 0.79

= - 457410 J

$\small \Delta G$⁰ = - RTlnK

or, - 457410 = - 8.314 $\small \times$ 298.15 ln K

or, ln K = 184.52

or, K = 1.4 $\small \times$10⁸⁰

cell reaction

Cr₂O₇^2- (aq)  + 14H⁺ (aq) + 6 I^- (aq)  $\small \to$ 2 Cr³⁺ (aq)  + 7H₂O (l) + 3 I₂ (s)

nernst equation of the reaction is

E_cell = E⁰_cell -  $\small \frac{0.0592}{6}$ log $\small \frac{[Cr^{3+}]^{2}}{[Cr_{2}O_{7}^{2-}][H^{+}]^{14}[I^{-}]^{6}}$

Or, E_cell = 0.79 - 0.00986 log $\small \frac{(4.28\times 10^{-2})^{2}}{(2.26\times 10^{-2})(2.31\times 10^{-3})^{14}[(3.55\times 10^{-3})^{6}}$

or, E_cell = 0.79 - 0.00986 log (3.29$\small \times$ 10⁵⁰)

or, E_cell = 0.79 - 0.00986$\small \times$ 50.52

or, Ecell = (0.79 - 0.50) V

or, E_cell = 0.29 V