Problem 20.51 Given the follow
Problem 20.51
Given the following reduction halfreactions:Fe3+(aq)+e??Fe2+(aq)E?red=+0.77VS2O2?6(aq)+4H+(aq)+2e??2H2SO3(aq)E?red=+0.60VN2O(g)+2H+(aq)+2e??N2(g)+H2O(l)E?red=?1.77VVO+2(aq)+2H+(aq)+e??VO2+(aq)+H2O(l)E?red=+1.00V
Part A
Write balanced chemical equation for the oxidation ofFe2+(aq) by S2O2?6 (aq).
Express your answer as a chemical equation. Identify all of thephases in your answer.


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Part B
Calculate ?G? for this reaction at 298 K.
Express your answer using two significant figures.


?G? =  kJ 
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Part C
Calculate the equilibrium constant K for this reactionat 298 K.
Express your answer using one significant figure.


K = 
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Part D
Write balanced chemical equation for the oxidation ofFe2+(aq) by N2O(g).
Express your answer as a chemical equation. Identify all of thephases in your answer.


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Part E
Calculate ?G? for this reaction at 298 K.
Express your answer using three significant figures.


?G? =  kJ 
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Part F
Calculate the equilibrium constant K for this reactionat 298 K.
Express your answer using one significant figure.


K = 
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Part G
Write balanced chemical equation for the oxidation ofFe2+(aq) by VO+2(aq).
Express your answer as a chemical equation. Identify all of thephases in your answer.


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Part H
Calculate ?G? for this reaction at 298 K.
Express your answer using two significant figures.


?G? =  kJ 
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Part I
Calculate the equilibrium constant K for this reactionat 298 K.
Express your answer using one significant figure.


K = 
Answer:
Part A :
Given chemical equations are :
Fe^{3+} (aq) + 1e^{–} ————>Fe^{2+} (aq) , E^{0}_{red.} = + 0.77 V………..(1)
S_{2}O_{6}^{2} (aq) +4H^{+}(aq) + 2e^{–} ———–> 2H_{2}SO_{3} (aq) , E^{0}_{red} = +0.60 V ……..(2)
multiply equation (1) by 2 and then reverse theequation , we have
2Fe^{2+} (aq) ————> 2Fe^{3+} (aq) +2e^{–} , E^{0}_{oxi} = – 0.77 V………..(3)
Now adding equation (2) and equation(3) , we have require balance equation is :
S_{2}O_{6}^{2} (aq) +4H^{+}(aq) + 2Fe^{2+} (aq)———> 2Fe^{3+} (aq) + 2H_{2}SO_{3} (aq)
also E^{0}_{cell} = E^{0}_{red}+ E^{0}_{ox}_{i} = 0.60 V – 0.77 V = – 0.17V .
Part B : Calculation of deltaG^{0} :
delta G^{0} = – n F E^{0}_{cell}
delta G^{0} = – 2 x 96500 C x – 0.17 V
delta G^{0} = 32810 J/mol
delta G^{0} = 32.81 KJ/mol
Part C : Calculation of k:
we know
delta G^{0} = – 2.303 RT log k
log k = – delta G^{0} / 2.303 RT
log k = – 32.81 KJ mol^{1} / 2.303 x 8.314 x10^{3} KJ K^{1} mol^{1} x 298 K
log k = – 5.75
k = 10^{5.75}
k = 1.78 x 10^{6}