Given the following reduction half-reactions: Fe^3+ (aq) + e^- rightarrow Fe^2+

Question

Given the following reduction half-reactions: Fe^3+ (aq) + e^- rightarrow Fe^2+ (aq) E_red = + 0.77V S_2O_6^2- (aq) + 2e^- rightarrow 2H_2SO_s (aq) E_red = + 0.60 V N_2O (g) + 2H^+ (aq) + 2e^- rightarrow N_2(g) + H_2O(l) E_red = - 1.77V VO_2 (aq) + 2H^+ (aq) + e^- rightarrow VO^2 + (aq) + H_2O(l) E_red = + 1.00V Write balanced chemical equation for the oxidation of Fe^2+ (aq) by S_2O_6^2- (aq). Express your answer as a chemical equation. Identify all of the phases in your answer. Calculate Delta G degree for this reaction at 298 K Express your answer using two significant figures. Calculate the equilibrium constant K for this reaction at 296 K Express your answer using one significant figure.

Explanation / Answer

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Remember that in a redox reaction, electrons are transferred from one substance to another.

Oxidation: Fe2+(aq) --> Fe3+(aq) + e-

Reduction: N2O(g) +2H+(aq) +2e- --> N2(g) +H2O(l)

The number of electrons in the oxidation reaction must be the same that the number of electrons in the reduction reaction. For that reason we multiply the oxidation reaction by 2

Oxidation: 2Fe2+(aq) --> 2Fe3+(aq) + 2e-

Reduction: N2O(g) +2H+(aq) +2e- --> N2(g) +H2O(l)

The electrons on both sides cancel, the balanced net ionic equation is

2Fe2+(aq) + N2O(g) +2H+(aq) -->  2Fe3+(aq) + N2(g) +H2O(l)

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