Q3.6: The cell in which the following reactions occurs:
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
Given that
n = 2
T = 298 K
Eocell = 0.236 V
We have the formula
∆rGθ = – nFEocell
Plug the values, we get
∆rGθ = −2 × 96487 × 0.236
∆rGθ = −45541.86 J mol−1
Divide by 100o to convert in KJ
∆rGθ = −45.54 kJ mol−1
Use second formula of ∆rGθ
∆rGθ =−2.303RT log Kc
Plug the values we get
−45541.86 J mol−1 = –2.303 × 8.314 × 298 log Kc
Solve it we get
Log Kc = 7.98
Take antilog both side, we get
Kc = Antilog (7.98)
= 9.6 × 107
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