Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
2Cr(s) + 3Cd²⁺ (aq) → 2Cr³⁺ (aq) + 3Cd
(Calculate the Δ_rG° and equilibrium constant of the reaction.)

Solution

E °_{{Cd}^{2 +} / Cd} = - 0.40 V, E °_{{Cr}^{3 +} / Cr} = - 0.74 V E °_{{Fe}^{3 +} / Fe} = 0.77 V, E °_{{Ag}^{+} / Ag} = + 0.80 V

Cr (s) {Cr}^{3 +} (aq) ∥ {Cd}^{2 +} (aq) Cd (s) (anode is on left and cathode on right) {E^{0}}_{cell} = {E^{0}}_{right} - {E^{0}}_{left} = - 0.40 - (0.74) = + 0.34 V

∆ {G_{r}}^{0} (or ∆ G °_{cell} = - nFE °_{cell}

(Here n = 6 (as 6e are involved in overall cell reaction i.e.,

2 Cr + 3 {Cd}^{2 +} \to 3 {Cr}^{3 +} + 3 Cd

)

= - 6 \times 96500 C \times 0.34 = - 196860 J {mol}^{- 1} = - 196.86 kJ {mol}^{- 1}

{E^{0}}_{cell} = \frac{0.0591}{n} \log K_{c} \log k_{c} = \frac{0.34 \times 6}{0.0591} = 34.5177

k_{c} = antilog 34.5177 = 3.294 \times 10^{34}

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