Problem 20.51 Given the follow

Problem 20.51

Given the following reduction half-reactions: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 :

Fe3+ (aq) + 1e ————->Fe2+ (aq) , E0red. = + 0.77 V………..(1)

S2O6-2 (aq) +4H+(aq) + 2e ———–> 2H2SO3 (aq) , E0red = +0.60 V ……..(2)

multiply equation (1) by 2 and then reverse theequation , 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) +   2H2SO3 (aq)

also E0cell = E0red+ E0oxi = 0.60 V – 0.77 V = – 0.17V .

Part B :- Calculation of deltaG0 :-

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 x10-3 KJ K-1 mol-1 x 298 K

log k = – 5.75

k = 10-5.75

k = 1.78 x 10-6


 
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