3.6 The cell in which the following reaction occurs: ( ) ( ) ( ) ( ) 3 2 2Fe 2I 2Fe I aq aqaq 2 s + − + + → + has 0 Ecell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction

Given: 
2Fe³⁺ (aq) + 2I^- (aq) → 2Fe²⁺ (aq) + I₂ (s)
E^0cell = 0.236V
n = moles of e^- from balanced redox reaction = 2
F = Faraday's constant = 96,485 C/mol
T = 298 K.
Using the formula, we get
? _rG⁰ = – nFE⁰_cell
⇒? _rG⁰ = – 2 × FE⁰_cell
⇒? _rG⁰ = −2 × 96485 C mol^-1 × 0.236 V
⇒? _rG⁰ = −45540 J mol^-1
⇒? _rG⁰ = −45.54 kJ mol^-1
Now,
? _rG⁰ = −2.303RT log K_c
Where, K is the equilibrium constant of the reaction

R is the gas constant; R = 8.314 J-mol-C^-1
⇒ −45540 J mol^-1 = –2.303× (8.314 J-mol-C^-1)× (298 K) × (log K_c)
Solving for K_c we get,
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Given: 
2Fe³⁺ (aq) + 2I^- (aq) → 2Fe²⁺ (aq) + I₂ (s)
E^0cell = 0.236V
n = moles of e^- from balanced redox reaction = 2
F = Faraday's constant = 96,485 C/mol
T = 298 K.
Using the formula, we get
? _rG⁰ = – nFE⁰_cell
⇒? _rG⁰ = – 2 × FE⁰_cell
⇒? _rG⁰ = −2 × 96485 C mol^-1 × 0.236 V
⇒? _rG⁰ = −45540 J mol^-1
⇒? _rG⁰ = −45.54 kJ mol^-1
Now,
? _rG⁰ = −2.303RT log K_c
Where, K is the equilibrium constant of the reaction

R is the gas constant; R = 8.314 J-mol-C^-1
⇒ −45540 J mol^-1 = –2.303× (8.314 J-mol-C^-1)× (298 K) × (log K_c)
Solving for K_c we get,
⇒ logK_c = 7.98
Taking antilog both side, we get
⇒ K_c = Antilog (7.98)
⇒ K_c = 9.6 × 107
Trick to remember

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<p>Given:&nbsp;<br>2Fe³⁺ (aq) + 2I^- (aq) → 2Fe²⁺ (aq) + I₂ (s)<br>E^0cell&nbsp;= 0.236V<br>n = moles of e^- from balanced redox reaction = 2<br>F = Faraday's constant = 96,485 C/mol<br>T = 298 K.<br>Using the formula, we get<br>? _rG⁰ = – nFE⁰_cell<br>⇒? _rG⁰ = – 2 × FE⁰_cell<br>⇒? _rG⁰ = −2 × 96485 C mol^-1 × 0.236 V<br>⇒? _rG⁰ = −45540 J mol^-1<br>⇒? _rG⁰ = −45.54 kJ mol^-1<br>Now, <br>? _rG⁰ = −2.303RT log K_c<br>Where, K is the equilibrium constant of the reaction</p><p>R is the gas constant; R = 8.314 J-mol-C^-1<br>⇒ −45540 J mol^-1 = –2.303× (8.314 J-mol-C^-1)× (298 K) × (log K_c)<br>Solving for K_c we get, <br>⇒ logK_c = 7.98<br>Taking antilog both side, we get<br>⇒ K_c = Antilog (7.98)<br>⇒ K_c = 9.6 × 107<br>Trick to remember</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686149073phpUc7n8i_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686149073phpUc7n8i.jpeg" alt="" width="266" height="126"></picture></div></div>

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