Given standard electrode potentials (Fe2+) + (2e-) ------------Fe, E= -0.440V (Fe3+) + (3e-) ------------Fe, E=-0.036V The standard electrode potential(E) for (Fe3+) + (e-) -------------(Fe2+) is (a) – 0.476V (b) – 0.404V (ć) +0.404V (d) +0.771V

Question

Given standard electrode potentials (Fe2+) + (2e-) ------------>Fe, E= -0.440V (Fe3+) + (3e-) ------------>Fe, E=-0.036V The standard electrode potential(E) for (Fe3+) + (e-) ------------->(Fe2+) is (a) – 0.476V (b) – 0.404V (ć) +0.404V (d) +0.771V

Wassim bhatt · Answer

Option d is the correct one. You cant add the potentials directly because EMF is the intensive property. So calculate the gibbs free energy values for all the three reactions and use G3 =G2-G1, you will get the value equal to 0.772v

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Given standard electrode potentials (Fe2+) + (2e-) ------------>Fe, E= -0.440V (Fe3+) + (3e-) ------------>Fe, E=-0.036V The standard electrode potential(E) for (Fe3+) + (e-) ------------->(Fe2+) is (a) – 0.476V (b) – 0.404V (ć) +0.404V (d) +0.771V

Given standard electrode potentials
(Fe2+) + (2e-) ------------>Fe, E= -0.440V
(Fe3+) + (3e-) ------------>Fe, E=-0.036V
The standard electrode potential(E) for
(Fe3+) + (e-) ------------->(Fe2+) is

(a) – 0.476V (b) – 0.404V
(ć) +0.404V (d) +0.771V

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Given standard electrode potentials (Fe2+) + (2e-) ------------>Fe, E= -0.440V (Fe3+) + (3e-) ------------>Fe, E=-0.036V The standard electrode potential(E) for (Fe3+) + (e-) ------------->(Fe2+) is (a) – 0.476V (b) – 0.404V (ć) +0.404V (d) +0.771V

Given standard electrode potentials (Fe2+) + (2e-) ------------>Fe, E= -0.440V(Fe3+) + (3e-) ------------>Fe, E=-0.036VThe standard electrode potential(E) for (Fe3+) + (e-) ------------->(Fe2+) is (a) – 0.476V (b) – 0.404V(ć) +0.404V (d) +0.771V

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Given standard electrode potentials
(Fe2+) + (2e-) ------------>Fe, E= -0.440V
(Fe3+) + (3e-) ------------>Fe, E=-0.036V
The standard electrode potential(E) for
(Fe3+) + (e-) ------------->(Fe2+) is

(a) – 0.476V (b) – 0.404V
(ć) +0.404V (d) +0.771V