Question 21.5: Writing Spontaneous Redox Reactions and Ranking Oxidizing an......
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
Problem (a) Combine pairs of the balanced half-reactions (1), (2), and (3) into three spontaneous reactions A, B, and C, and calculate E^°_{\text{cell}} for each. (b) Rank the relative strengths of the oxidizing and reducing agents.
(1) NO_3^−(aq) + 4H^+(aq) + 3e^− ⟶ NO(g) + 2H_2O(l) E^\circ = 0.96 V
(2) N_2(g) + 5H^+(aq) + 4e^− ⟶ N_2H_5^+(aq) E^\circ = −0.23 V
(3) MnO_2(s) + 4H^+(aq) + 2e^− ⟶ Mn^{2+}(aq) + 2H_2O(l) E^\circ = 1.23 V
Plan (a) To write spontaneous redox reactions, we combine the possible pairs of half-reactions: (1) and (2) give reaction A, (1) and (3) give B, and (2) and (3) give C. They are all written as reductions, so the oxidizing agents appear as reactants and the reducing agents appear as products. For each pair, we reverse the half-reaction that has the smaller (less positive or more negative) E° value to an oxidation (without changing the sign of E°) to obtain a positive E^\circ _{\text{cell}}. We make e^− lost equal e^− gained (without changing the E° value), add the half-reactions together, and then apply Equation 21.3 to find E^\circ _{\text{cell}}. (b) Because each reaction is spontaneous as written, the stronger oxidizing and reducing agents are the reactants. To obtain the overall ranking, we first rank the relative strengths within each equation and then compare them.
E^\circ _{\text{cell}} = E^\circ _{\text{cathode(reduction)}} − E^\circ _{\text{anode(oxidation)}} (21.3)
Solution (a) Combining half-reactions (1) and (2) gives equation A. The E° value for half-reaction (1) is larger (more positive) than that for (2), so we reverse (2) to obtain a positive E^\circ_{\text{cell}}:
(1) NO_3^−(aq) + 4H^+(aq) + 3e^− ⟶ NO(g) + 2H_2O(l) E^\circ = 0.96 V
(rev 2) N_2H_5^+(aq) ⟶ N_2(g) + 5H^+(aq) + 4e^− E^\circ = −0.23 V
To make e^− lost equal e^− gained, we multiply (1) by four and the reversed (2) by three; then add half-reactions and cancel appropriate numbers of common species (H^+ and e^−):
4NO_3^−(aq) + 16H^+(aq) + 12e^− ⟶ 4NO(g) + 8H_2O(l) E^\circ = 0.96 V
3N_2H_5^+(aq) ⟶ 3N_2(g) + 15H^+(aq) + 12e^− E^\circ = −0.23 V
\overline{(A) 3N_2H_5^+(aq) + 4NO_3^−(aq) + H^+(aq) ⟶ 3N_2(g) + 4NO(g) + 8H_2O(l)}
E^\circ _{\text{cell}} = 0.96 V − (−0.23 V) = 1.19 V
Combining half-reactions (1) and (3) gives equation B. Half-reaction (1) has a smaller E°, so it is reversed:
(rev 1) NO(g) + 2H_2O(l) ⟶ NO_3^−(aq) + 4H^+(aq) + 3e^− E^\circ = 0.96 V
(3) MnO_2(s) + 4H^+(aq) + 2e^− ⟶ Mn^{2+} (aq) + 2H_2O(l) E^\circ = 1.23 V
We multiply reversed (1) by two and (3) by three, then add and cancel:
2NO(g) + 4H_2O(l) ⟶ 2NO_3^−(aq) + 8H^+(aq) + 6e^− E^\circ = 0.96 V
3MnO_2(s) + 12H^+(aq) + 6e^− ⟶ 3Mn^{2+}(aq) + 6H_2O(l) E^\circ = 1.23 V
\overline{(B) 3MnO_2(s) + 4H^+(aq) + 2NO(g) ⟶ 3Mn^{2+}(aq) + 2H_2O(l) + 2NO_3^−(aq) }
E^\circ _{\text{cell}} = 1.23 V − 0.96 V = 0.27 V
Combining half-reactions (2) and (3) gives equation C. Half-reaction (2) has a smaller E°, so it is reversed:
(rev 2) N_2H_5^+(aq) ⟶ N_2(g) + 5H^+(aq) + 4e^− E^\circ = −0.23 V
(3) MnO_2(s) + 4H^+(aq) + 2e^− ⟶ Mn^{2+}(aq) + 2H_2O(l) E^\circ = 1.23 V
We multiply reaction (3) by two, add the half-reactions, and cancel:
N_2H_5^+(aq) ⟶ N_2(g) + 5H^+(aq) + 4e^− E^\circ = −0.23 V
2MnO_2(s) + 8H^+(aq) + 4e^− ⟶ 2Mn^{2+}(aq) + 4H_2O(l) E^\circ = 1.23 V
\overline{(C) N_2H_5^+(aq) + 2MnO_2(s) + 3H^+(aq) ⟶ N_2(g) + 2Mn^{2+}(aq) + 4H_2O(l)}
E^\circ _{\text{cell}} = 1.23 V − (−0.23 V) = 1.46 V
(b) Ranking the oxidizing and reducing agents within each reaction:
Equation A Oxidizing agents: NO_3^− > N_2 Reducing agents: N_2H_5^+ > NO
Equation B Oxidizing agents: MnO_2 > NO_3^− Reducing agents: NO > Mn^{2+}
Equation C Oxidizing agents: MnO_2 > N_2 Reducing agents: N_2H_5^+ > Mn^{2+}
Determining the overall ranking of oxidizing and reducing agents. Comparing the
relative strengths from the three balanced equations gives
\text{Oxidizing agents: }MnO_2 > NO_3^− > N_2
\text{Reducing agents: }N_2H_5^+ > NO > Mn^{2+}
Check As always, check that atoms and charges balance on both sides of each equation. A good way to check the ranking and equations is to list the given half-reactions in order of decreasing E° value:
MnO_2(s) + 4H^+(aq) + 2e^− ⟶ Mn^{2+}(aq) + 2H_2O(l) E^\circ = 1.23 V
NO_3^−(aq) + 4H^+(aq) + 3e^− ⟶ NO(g) + 2H_2O(l) E^\circ = 0.96 V
N_2(g) + 5H^+(aq) + 4e^− ⟶ N_2H_5^+(aq) E^\circ = −0.23 V
Then the oxidizing agents (reactants) decrease in strength going down the list, so the reducing agents (products) decrease in strength going up. Moreover, each of the equations for the spontaneous reactions (A, B, and C) should combine a reactant with a product that is lower down on this list.