In: Chemistry
Problem 20.51
Given the following reduction half-reactions:
Fe3+(aq)+e??Fe2+(aq)
E?red=+0.77V
S2O2?6(aq)+4H+(aq)+2e??2H2SO3(aq)
E?red=+0.60V
N2O(g)+2H+(aq)+2e??N2(g)+H2O(l)
E?red=?1.77V
VO+2(aq)+2H+(aq)+e??VO2+(aq)+H2O(l)
E?red=+1.00V
Part A
Write balanced chemical equation for the oxidation of Fe2+(aq) by S2O2?6 (aq).
Express your answer as a chemical equation. Identify all of the phases in your answer.
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Part B
Calculate ?G? for this reaction at 298 K.
Express your answer using two significant figures.
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?G? = | kJ |
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Part C
Calculate the equilibrium constant K for this reaction at 298 K.
Express your answer using one significant figure.
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K = |
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Part D
Write balanced chemical equation for the oxidation of Fe2+(aq) by N2O(g).
Express your answer as a chemical equation. Identify all of the phases in your answer.
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Part E
Calculate ?G? for this reaction at 298 K.
Express your answer using three significant figures.
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?G? = | kJ |
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Part F
Calculate the equilibrium constant K for this reaction at 298 K.
Express your answer using one significant figure.
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K = |
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Part G
Write balanced chemical equation for the oxidation of Fe2+(aq) by VO+2(aq).
Express your answer as a chemical equation. Identify all of the phases in your answer.
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Part H
Calculate ?G? for this reaction at 298 K.
Express your answer using two significant figures.
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?G? = | kJ |
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Part I
Calculate the equilibrium constant K for this reaction at 298 K.
Express your answer using one significant figure.
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K = |
Part A :-
Given chemical equations are :
Fe3+ (aq) + 1e- -------------> Fe2+ (aq) , E0red. = + 0.77 V ...........(1)
S2O6-2 (aq) + 4H+(aq) + 2e- -----------> 2 H2SO3 (aq) , E0red = + 0.60 V ........(2)
multiply equation (1) by 2 and then reverse the equation , we have
2Fe2+ (aq) ------------> 2Fe3+ (aq) + 2e- , E0oxi = - 0.77 V ...........(3)
Now adding equation (2) and equation (3) , we have require balance equation is :
S2O6-2 (aq) + 4H+(aq) + 2Fe2+ (aq) ----------> 2Fe3+ (aq) + 2 H2SO3 (aq)
also E0cell = E0red + E0oxi = 0.60 V - 0.77 V = - 0.17 V .
Part B :- Calculation of delta G0 :-
delta G0 = - n F E0cell
delta G0 = - 2 x 96500 C x - 0.17 V
delta G0 = 32810 J/mol
delta G0 = 32.81 KJ/mol
Part C :- Calculation of k :-
we know
delta G0 = - 2.303 RT log k
log k = - delta G0 / 2.303 RT
log k = - 32.81 KJ mol-1 / 2.303 x 8.314 x 10-3 KJ K-1 mol-1 x 298 K
log k = - 5.75
k = 10-5.75
k = 1.78 x 10-6