Question

how to predict the ions present in an aqueous solution.

it is a reaction between (KMnO4) and (H2C2O4).

Answer 1

This involves redox reaction. Oxalic acid is oxidised to CO₂ by KMnO₄ (Mn⁷⁺) which itself gets reduced to Mn²⁺. The reaction may takes place in acidic medium or basic medium.

In acidic medium the net balanced equation is-

2MnO⁴⁻+ 5H₂C₂O₄ + 6H⁺   ⟶ 2Mn²⁺  + 10CO₂  + 8H₂O

In Basic medium-

2MnO⁴⁻+ 3H₂C₂O₄ + H₂O + 8OH^-   ⟶ 2OH^- + 2MnO₂  + 6CO₂  + 11H₂O

Balancing of ions is given below-

Consider the reaction between permanganate ion and oxalic acid to form manganese (II) ion and carbon dioxide in acidic medium. The overall balanced redox equation can be formed using following steps-

Write the skeletal equation for the reaction; identify what is oxidized and reduced. C is oxidized, +3 → +4 H-C-04 is the reducing agent Mn is reduced, +7 -+2 .. MnO4 is the oxidizing agent Write the skeletal half reactions for the oxidation and reduction. oxidation: H,C,04 → reduction: MnO4 Mn CO2 Balance elements other than H and O oxidation: H2C204 -> reduction: MnO -→ Mn2+ Balanced O by adding H2O oxidation: HaCO reduction: MnO → Mn2++ 4 HO 2 Balance H by adding H'. oxidation: HC204 reduction: 8H* + MnO4 -Mn 2 CO2 2 H 4 H2O Balance charge by adding electrons (e) oxidation: HCo2 CO, 2 H +2 e reduction: 5e + 8 H* + MnO4 Mn*+ 4 H2O

Prepare to add the half reacitons together by multiplying by an appropriate factor so that the electrons cancel out. 5 x [ oxidation: H-C,O, 2 CO, + 2 H+ + 2 e-l 2 x [reduction : 5 e_ + 8 H+ + MnO,-→ Mn2++ 4 H,0 I oxidation: 5 H,C,04 → 10 CO, + 10 H+ + 10 e- ducion: 10e 16 Hr + 2 MnO, 2 Mn+ 8 HO Add 1/2 reactions together, simplify (cancel species that are the same on both sides of the equation) ; add physical states cancel e, reduce H 5 H,C,04 (aq) + 6 H, (aq) + 2 MnO,-(aq) → 10 CO, (g) + 2 Mn2+ (aq) + 8 H,O φ