redox balance. H3MO3 + MnO4^- → H3MO4 + Mn^2+ (acidic)

redox balance. H3MO3 + MnO4^- → H3MO4 + Mn^2+ (acidic)

산화-환원 반응 완성하기. MnO4^- + H3MO3 → H3MO4 + Mn^2+

H3MO3(aq) + MnO4^-(aq) → H3MO4(aq) + Mn^2+(aq) (in acid)

 

 

On heating a 0.200 g sample of a certain semimetal in air, the corresponding oxide M2O3 was obtained. When the oxide was dissolved in aqueous acid and titrated with KMnO4, 10.7 mL of 0.100 M MnO4^- was required for complete reaction. The unbalanced equation is
H3MO3(aq) + MnO4^-(aq) → H3MO4(aq) + Mn^2+(aq) (in acid)
(a) Balance the equation.

 

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▶ 산화-환원 반응 균형 맞추기 설명 (산성 조건)

https://ywpop.tistory.com/4264

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1. 반쪽 반응식 나누기

H3MO3 → H3MO4 (M의 산화수는 +3에서 +5로 증가, H3MO3는 산화됨.)

MnO4^- → Mn^2+ (Mn의 산화수는 +7에서 +2로 감소, MnO4^-는 환원됨.)

 

 

2. 질량 균형 맞추기

H3MO3 + H2O → H3MO4 + 2H^+

MnO4^- + 8H^+ → Mn^2+ + 4H2O

 

 

3. 전하 균형 맞추기

H3MO3 + H2O → H3MO4 + 2H^+ + 2e^-

MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O

 

 

4. 주고받은(이동한) 전자 수 같도록

H3MO3 + H2O → H3MO4 + 2H^+ + 2e^- (×5)

MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O (×2)

 

5H3MO3 + 5H2O → 5H3MO4 + 10H^+ + 10e^-

2MnO4^- + 16H^+ + 10e^- → 2Mn^2+ + 8H2O

 

 

5. 반쪽 반응 더하기. 끝.

5H3MO3 + 2MnO4^- + 6H^+ → 5H3MO4 + 2Mn^2+ + 3H2O

5H3MO3(aq) + 2MnO4^-(aq) + 6H^+(aq) → 5H3MO4(aq) + 2Mn^2+(aq) + 3H2O(l)

 

 

 

[ 관련 예제 https://ywpop.tistory.com/11694 ]

 

 

 

[키워드] redox H3MO3, redox MnO4^-