Calculate the equilibrium constant for the reaction- Fe2- - Ce4- Ce3- - Fe3-Given- ECe4-Ce3- - 1-44 V and EFe3-Fe2- - 0-68 V-Take 2-303 RTF-0-06 at 25oC- log 4-68-0-67

E^∘_cell = 0.0591log K_c E^∘_cell = E^∘_Fe^2+/Fe^3+ + E^∘_Ce^4+/Ce^3+ = 0.68 + 1.44 = 0.76 V log_10K_c = 0.760.059 = 12.8814 K_c = 7.6 ...

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Equilibrium Constant from Nernst Equation

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Question

Calculate the equilibrium constant for the reaction,
$F e^{2 +} + C e^{4 +} ⇌ C e^{3 +} + F e^{3 +}$
Given, $E_{C e^{4 +} / C e^{3 +}}^{\circ} = 1.44 V$ and $E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.68 V$ .
Take $\frac{2.303 R T}{F} = 0.06$ at $25^{o} C, l o g 4.68 = 0.67$

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F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} | C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} | F e^{2 +}}^{o} = 0.68 V

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} / C e^{3 +}}^{0} = 1.44 V; E_{F e^{3 +} / F e^{2 +}}^{0} = 0.68 V]

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{{C e}^{4 +} / C e^{3 +}}^{\circ} = 1.44 V; E_{F e^{2 +} / F e^{3 +}}^{\circ} = 0.68 V

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

E_{C e^{4 +} / C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} / F e^{2 +}}^{o} = 0.68 V

F e^{2 +} + C e^{4 +} | F e^{3 +} + C e^{3 +}

E^{0} C e^{4 +} / C e^{3 +} = 1.44

E^{0} F e^{3 +} / F e^{2 +} = 0.68 V]

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