Question

Calculate the value of $ΔG$ and $E_{cell}$ at $298 K$ for the following cell.

$Al |{AI}_{(0.01 M)}^{3 +}| |{Fe}_{(0.02 M)}^{2 +}| Fe$

Given that $E_{{Al}^{3 +} / Al}^{Θ} = - 1.66 V$ and $E_{{Fe}^{2 +} / Fe}^{Θ} = 0.80 V$

Accepted Answer

To Calculate the value of $ΔG$ and $E_{cell}$ at $298 K$ for the following cell we will make use of Nernst equation

$Al |{AI}_{(0.01 M)}^{3 +}| |{Fe}_{(0.02 M)}^{2 +}| Fe The above cell representation has electrode where electron transfer reaction occurs Al |{AI}_{(0.01 M)}^{3 +}| as anode, oxidation occurs [Al \to {Al}^{3 +} + 3 e^{-}] x 2, |{Fe}_{(0.02 M)}^{2 +}| Fe as cathode, reduction occurs [{Fe}^{2 +} + 2 e^{-} \to Fe] x 3 The overall cell reaction 2 Al + 3 {Fe}^{2 +} \to 2 {Al}^{3 +} + 3 Fe, n = 6 e^{-}$

Given that $E_{{Al}^{3 +} / Al}^{Θ} = - 1.66 V and E_{{Fe}^{2 +} / Fe}^{Θ} = + 0.80 V$ , $using the electrode potential values we find the {E^{°}}_{cell} {E^{°}}_{cell} = {E^{°}}_{right} - {E^{°}}_{left} = E^{°}_{cathode} - {E^{°}}_{anode} {E^{°}}_{cell} = E_{{Fe}^{2 +} / Fe}^{Θ} - E_{{Al}^{3 +} / Al}^{Θ} {E^{°}}_{cell} = 0.80 V - (- 1.66 V) {E^{°}}_{cell} = 1.2308 V$

$To find the Change in Gibb' s Free Energy we make use of the formula with Nernst Equation E_{cell} = {E^{°}}_{cell} - \frac{0.059}{n} \log \frac{[products]}{[reactants]} E_{cell} = {E^{°}}_{cell} - \frac{0.059}{n} \log \frac{[{Al}^{3 +}]}{[{Fe}^{2 +}]}, Taking unit concentration of metal Al, Fe E_{cell} = 1.2308 - \frac{0.059}{6} \log \frac{{[0.01]}^{2}}{{[0.02]}^{3}} E_{cell} = 1.2308 - \frac{0.059}{6} \log \frac{{[0.01]}^{2}}{{[0.02]}^{3}} E_{cell} = 1.2308 - \frac{0.059}{6} \log \frac{[0.0001]}{[0.0008]} E_{cell} = 1.2308 - \frac{0.059}{6} \log \frac{1}{8} E_{cell} = 1.2308 - \frac{0.059}{6} (- 0.903) E_{cell} = 1.2308 + \frac{0.059}{6} (0.903) E_{cell} = 1.2396 V$

The next step is using the value of emf of cell, find the value of GIbb's Free Energy Change. $The Cell Potential as EMF is related to Gibb' s Free Energy as ∆ G = - {nFE}^{°}_{cell} ∆ G = - 9 x 96500 x 1.2396 ΔG = - 71777.4 J ΔG = - 712.633 kJ$

K L Chugh Solutions for Chapter: Electrochemistry, Exercise 10: Follow Up Problems

K L Chugh Chemistry Solutions for Exercise - K L Chugh Solutions for Chapter: Electrochemistry, Exercise 10: Follow Up Problems

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