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albertruru
@albertruru
August 2022
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298. Составьте уравнения окислительно-восстановительных реакций методом
полуреакций:
1) SnSO4 + К2Сr2О7 + H2SO4 → Sn(SO4)2 +Cr2(SO4)3 + K2SO4 + H2O;
2) Р + Н2SО4(конц.) → Н3РО4 + SO2 + Н2О;
3) МnО2 + КСlO3 + КОН → K2MnO4 + KC1 + Н2О.
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Answers & Comments
wasurenagusa
1) 3SnSO4 + К2Сr2О7 + 7H2SO4 = 3Sn(SO4)2 + Cr2(SO4)3 + K2SO4 + 7H2O;
Sn²⁺ - 2e = Sn⁴⁺ x3
Cr2O7²⁻ + 6e + 14H⁺ = 2Cr³⁺ + 7H2O
2) 2Р + 5Н2SО4(конц.) = 2Н3РО4 + 5SO2 + 2Н2О;
P⁰ - 5e + 4H2O = PO4³⁻ + 8H⁺ x2
SO4²⁻ + 2e + 4H⁺ = SO2 + 2H2O x5
3) 3МnО2 + КСlO3 + 6КОН = 3K2MnO4 + KCl + 3Н2О.
MnO2 - 2e + 4OH
⁻ = MnO4²⁻ + 2H2O x3
ClO3
⁻ + 6e + 3H2O
= Cl⁻ + 6OH⁻
2 votes
Thanks 1
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Answers & Comments
Sn²⁺ - 2e = Sn⁴⁺ x3
Cr2O7²⁻ + 6e + 14H⁺ = 2Cr³⁺ + 7H2O
2) 2Р + 5Н2SО4(конц.) = 2Н3РО4 + 5SO2 + 2Н2О;
P⁰ - 5e + 4H2O = PO4³⁻ + 8H⁺ x2
SO4²⁻ + 2e + 4H⁺ = SO2 + 2H2O x5
3) 3МnО2 + КСlO3 + 6КОН = 3K2MnO4 + KCl + 3Н2О.
MnO2 - 2e + 4OH⁻ = MnO4²⁻ + 2H2O x3
ClO3⁻ + 6e + 3H2O = Cl⁻ + 6OH⁻