Problem 20.51 Given the following reduction half-reactions: Fe3+( a q )+e??Fe2+(

Question

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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Part C

Calculate the equilibrium constant K for this reaction at 298 K.

Express your answer using one significant figure.

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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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Part F

Calculate the equilibrium constant K for this reaction at 298 K.

Express your answer using one significant figure.

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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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Part I

Calculate the equilibrium constant K for this reaction at 298 K.

Express your answer using one significant figure.

Explanation / Answer

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

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