Question

Given \( E_{F e^{+3} / F^{+2}}^{0}=+0.76 V \) and \( E_{l_{2} / r}^{0}=+0.55 V \). The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is \( \left[\frac{2.303 R T}{F}=0.06\right] \) (2020) a) \( 1 \times 10^{7} \) b) \( 1 \times 10^{9} \) c) \( 3 \times 10^{8} \) d) \( 5 \times 10^{12} \)

Answer 1

Given, \(E^0_{Fe^{+3}/Fe^{+2}}\)​ = +0.76V (Cathode)

\(E_{I_2/I^-}\)​ = +0.55V (Anode)

∴ \(E_{Cell}^0 = E^0_C - E^0_A\) ​= 0.76 − 0.55 = 0.21

The complete cell reaction is

\(2Fe^{3+} + 2I^- \to 2Fe^{2+} +I_2\)

∴ \(E_{Cell}^0 = \frac{0.059}2 \log K_C\)

⇒ logK_C​ ≈ 7

⇒ K_C ​= 10⁷