Determine the equilibrium constant, Keq, at 25°C for the reaction 2Br– (aq) + I2(s) Br2(l) + 2I– (aq)
5.7 ? 10–19 is the correct answer
Can someone explain how they got this answer
The reaction
2Br - (aq) + I2 (s) Br2(l) + 2I - (aq)
The standard reduction potentials of the half-cell reactions are followings
Br2 (l) + 2e- 2Br-(aq) E0 = +1.07V
I2(s) + 2e- 2I -(aq) E°= +0.53V
Overall
2Br - (aq) + I2 (s) Br2(l) + 2I - (aq)
We know,
E0cell = E0cathode
- E0anode
E0cell = +0.53V - 1.07V
E0cell = -0.54V
We know
E0cell = (0.0592/n) log K (K is the equilibrium constant)
log K = (-0.54*2)/(0.0592)
K = 5.7 *10-19(Answer)
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