Answer:
6H⁺ + 2MnO₄⁻ + 5H₂C₂O₄ → 2Mn²⁺ + 10CO₂ + 8H₂O
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄(aq) → Mn²⁺(aq) + CO₂(g)
First of all, determinate the half reaction (reduction/oxidation)
Reduction - Oxidation state decrease
Oxidation - Oxidation state increase
MnO₄⁻ → Mn²⁺
Manganese acts with +7 in permanganate, and it's get reduced to Mn²⁺
C₂O₄⁻² → CO₂
Carbon changes from +3 to +4. It's been oxidized.
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
We add 4 water in products side because we have 4 oxygen in reactant side (acidic medium) . Aftewards, we balance the protons. Mn to change from +7 to +2, has gained 5 moles of electrons.
C₂O₄⁻² → 2CO₂ + 2e⁻
We can balance the carbon with 2, so now the half reaction is balanced in charges and atoms. Carbon has released 2 mol of e⁻ (3+ → 4+)
(8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O) .2
( C₂O₄⁻² → 2CO₂ + 2e⁻) .5
Let's multiply the reaction x2 and x5, to balance the electrons
16H⁺ + 2MnO₄⁻ + 10e⁻ → 2Mn²⁺ + 8H₂O
5C₂O₄⁻² → 10CO₂ + 10e⁻
We sum both reaction, so we can quit the electrons
16H⁺ + 2MnO₄⁻ + 5C₂O₄⁻² → 2Mn²⁺ + 10CO₂ + 8H₂O
Finally the balanced redox reaction is this:
6H⁺ + 2MnO₄⁻ + 5H₂C₂O₄ → 2Mn²⁺ + 10CO₂ + 8H₂O
Redox reaction equations involve loss and gain of electrons
To balance the redox reaction, we must first break the whole reaction into half equations;
Oxidation half equation;
5C2O4^2-(aq) + 10e + 8H^+-----> 10CO2(g) + 4H2O(l)
Reduction half equation;
2MnO4^- (aq) + 10e + 8H^+-----> 2Mn^2+(aq) + 4H2O(l)
Having written the two half equation, we can now combine, excluding the number of electrons lost or gained to obtain;
5C2O4^2-(aq) + 2MnO4^- (aq) + 16H^+-----> 10CO2(g) + 2Mn^2+(aq) + 8H2O(l)
The coefficient of H2C2O4 is 5 and that of water is 8
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