Answer:
MnO4 − was reduced
C2 O4^2- was oxidized
Explanation:
2MnO4 −(aq) +5H +(aq) +2 C2 O4^2-(aq) +6H+(aq) →2Mn2+(aq) + 10CO2(g) +8H2 O(l)
A redox reaction is characterized by a change in the oxidation number of reacting species from left to right in the reaction equation.
If we look at Manganese, its oxidation number was decreased from +7 on the left hand side to +2 on the right hand side. This implies that it gained five electrons. recall that reduction refers to electron gain.
If we look at carbon, its oxidation number increased from +3 on the left hand side to +4 on the right hand side. This indicates loss of one electron and oxidation refers to electron loss.
MnO₄⁻ was reduced and C₂O₄²⁻ was oxidized in the titration.
The balanced chemical equation for the reaction can be expressed as:
[tex]\mathbf{6H^+_{(aq0} + 2MnO_4^-_{(aq)} + 5H_2C_2O_4 _{(aq)} \to 10CO_{2(g)} +8H_2O_{(l)} + 2Mn^{2+}_{(aq)}}[/tex]
A redox reaction can be described by a variation in the oxidation number of reacting species in the chemical reaction equation from the reactant side to the product side
Let consider Mn in MnO₄⁻, its oxidation number was decreased from +7 on the reactant side to +2 on the product side. This shows a gain of additional five electrons.
Thus, MnO₄⁻ is reduced.
Also in C₂O₄²⁻, the oxidation number of carbon increases from +3 on the reactant side to +4 on the product side. Since an electron is lost, then oxidation occurs at C₂O₄²⁻
Learn more about oxidation and reduction here:
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