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Zorluk: OrtaBalancing Redox Equations and Half-Reactions
Consider the redox reaction taking place in an acidic medium:
MnO4−(aq)+H2O2(aq)+H+(aq)→Mn2+(aq)+O2(g)+H2O(l)\text{MnO}_4^-(\text{aq}) + \text{H}_2\text{O}_2(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{O}_2(\text{g}) + \text{H}_2\text{O}(\text{l})
When this chemical equation is balanced using the smallest whole-number coefficients, what is the stoichiometric coefficient of H+(aq)\text{H}^+(\text{aq})?
  1. A
    16
  2. 6Cevap
  3. C
    8
  4. D
    3

Cevap

6
The balanced net redox equation is 2MnO4−(aq)+5H2O2(aq)+6H+(aq)→2Mn2+(aq)+5O2(g)+8H2O(l)2\text{MnO}_4^-(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{O}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l}). Thus, the stoichiometric coefficient of H+(aq)\text{H}^+(\text{aq}) is 6.

Adım Adım Çözüm

1
Write and balance the reduction half-reaction
MnO4−+8H++5e−→Mn2++4H2O\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
Manganese is reduced from oxidation state +7 in MnO4−\text{MnO}_4^- to +2 in Mn2+\text{Mn}^{2+}, requiring 5 electrons. Charge and mass are balanced using H+\text{H}^+ and H2O\text{H}_2\text{O}.
2
Write and balance the oxidation half-reaction
H2O2→O2+2H++2e−\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-
Oxygen in H2O2\text{H}_2\text{O}_2 is oxidized from -1 to 0 in O2\text{O}_2, releasing 2 electrons per molecule.
3
Equalize electron loss and gain
Multiply reduction half-reaction by 2 and oxidation half-reaction by 5:
2MnO4−+16H++10e−→2Mn2++8H2O2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}
5H2O2→5O2+10H++10e−5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10\text{e}^-
The least common multiple of 5 and 2 electrons transferred is 10.
4
Combine the half-reactions and simplify redundant species
2MnO4−+5H2O2+6H+→2Mn2++5O2+8H2O2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}
Subtracting 10H+10\text{H}^+ and 10e−10\text{e}^- from both sides yields a net coefficient of 6 for H+\text{H}^+ on the reactant side.

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