1) The free energy change for the following reaction at 25 °C,
when [Pb^{2+}] = 1.18 M
and [Cd^{2+}] =
7.90×10^{-3} M, is -65.9
kJ:
Pb^{2+}(1.18 M) +
Cd(s)>
Pb(s) +
Cd^{2+}(7.90×10^{-3}
M) ΔG = -65.9 kJ
What is the cell potential for the reaction as written under these
conditions?
Answer: ___V
Would this reaction be spontaneous in the forward or the reverse
direction?
2) Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction:
2Cr^{3+}(aq) + Zn(s)>2Cr^{2+}(aq) + Zn^{2+}(aq)
Answer: ___ kJ
K for this reaction would be __greater or less__ than one.
3) Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction:
2Fe^{3+}(aq) + Fe(s) > 2Fe^{2+}(aq) + Fe^{2+}(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values for physical constants.
Equilibrium constant: ___
DeltaG° for this reaction would be __greater or less__ than zero.
4) Consider the following reaction at 298 K.
2
Cr^{3+}(aq) +
Zn(s) > 2
Cr^{2+}(aq) +
Zn^{2+}(aq)
Which of the following statements are correct?
Choose all that apply.
a) delta G^{o} > 0
b) n = 2 mol electrons
c) E^{o}_{cell} > 0
d) K < 1
e) The reaction is product-favored.
1.Ans :- E_{cell} = 0.34 V (Reaction is Spontaneous in forward direction).
Explanation :-
As,
∆G = ∆G° + 2.303 RT log Q
Here, R= Universal gas constant = 8.314 x 10^{-3} KJ K^{-1}mol^{-1}
T= Temperature = 298 K
Q = Reaction quotient = [Products]^{stoichiometric coefficient}/[Reactants]^{stoichiometric coefficient}
So,
-65.9 KJ = ∆G° + 2.303 x 8.314 x 10^{-3} KJ K^{-1}mol^{-1} x 298 K log [Cd^{2+}]/[Pb^{2+}]
-65.9 KJ = ∆G° + 2.303 x 8.314 x 10^{-3} KJ K^{-1}mol^{-1} x 298 K log(7.90×10^{-3} M) /(1.18 M)
-65.9 KJ = ∆G° - 12.41 KJ
So,
∆G° = - 53.5 KJ
Also,
∆G° = -nFE°_{cell}
E°_{cell}= - ∆G°/nF
= - (-53.5 x 10^{3} J) /(2).(96500 C)
E°_{cell} = 0.277 V
By using Nernst equation :
= E°_{cell} - 0.0591/n. logQ
= 0.277 V - 0.0591/2 . log(7.90×10^{-3} M) /(1.18 M)
= 0.277 V + 0.0642
= 0.34 V
Hence, cell potential E_{cell} = 0.34 V Also, This reaction be spontaneous in the forward direction as E_{cell} = +ve. |
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