Reaction E°, V Mg 2+ (aq) + 2e – Mg (s) -2.356 Fe 2+ (aq) + 2e – Fe (s) -0.440 P
ID: 523618 • Letter: R
Question
Reaction E°, V
Mg2+ (aq) + 2e– Mg (s) -2.356
Fe2+ (aq) + 2e– Fe (s) -0.440
Pb2+ (aq) + 2e– Pb (s) -0.125
2 H+ (aq) + 2e– H2 (g) 0.0
Cu2+ (aq) + 2e– Cu (s) +0.337
I2 (s) + 2e– 2 I– (s) +0.535
Fe3+ (aq) + 2e– Fe2+ (aq) +0.771
Ag+ (aq) + e– Ag (s) +0.800
O2 (g) + 4 H+ (aq) + 2e– 2 H2O +1.229
Cl2 (g) + 2e– 2 Cl– (aq) +1.358
F2 (g) + 2e– 2 F– (aq) +2.866
Calculate the value of Keq at 25°C for the reaction:
F2 (g) + 2 I– 2 F– (aq) + I2 (g)
A) 6.0 × 1078
B) 1.4 × 10–79
C) 2.33
D) 2.6 × 1039
E) 5.3 × 1039
Please Explain!!
Explanation / Answer
from data table:
Eo(I2(s)/2I-) = 0.535 V
Eo(F2(g)/2F-) = 2.866 V
the electrode with the greater Eo value will be reduced and it will be cathode
here:
cathode is (F2(g)/2F-)
anode is (I2(s)/2I-)
The chemical reaction taking place is
F2(g) + 2I- --> 2F- + I2(s)
Eocell = Eocathode - Eoanode
= (2.866) - (0.535)
= 2.331 V
here, number of electrons being transferred, n = 2
E = (2.303*R*T)/(n*F) log Kc
At 25 oC or 298 K, R*T/F = 0.0592
So, E = (0.0592/n)*log Kc
2.331 = (0.0592/2)*log Kc
log Kc = 78.75
Kc = 6.0*10^78
Answer: A
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