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RC MUKHERJEE-OXIDATION NUMBER AND BALANCING OF REDOX REACTIONS -PROBLEMS
- H(2) O(2) + ClO(2) + OH^(-) = ClO(2)^(2-) + OH^(-)
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- ClO^(-) + CrO(2)^(2-) + OH^(-) = Cl^(-) + CrO(4)^(2-) + H(2)O
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- I(2) + Cl(2) + H(2) O = HIO(3) + HCl
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- Cl(2) + KOH = KOCl + KCl + H(2)O
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- Cl(2) + KOH = KClO(3) + KCl + H(2) O
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- H(2) O(2) + I(2) = HIO(3) + H(2)O
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- H(2) O(2) + KMnO(4) = MnO(2) + KOH + O(2) + H(2)O
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- HNO(2) + KMnO(4) + H(2) SO(4) = HNO(3) + KMnO(4) + K(2) SO(4) + H(2)O
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- NaNO(2) + NaI + H(2) SO(4) = NO + I(2) + Na(2) SO(4) + H(2)O
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- N(2) H(4) + AgNO(3) + KOH = N(2) + Ag + KNO(3) + H(2) O
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- N(2) H(4) + Zn + KOH + H(2) O = NH(3) + K(2) [ Zn (OH)(4)]
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- Fe + N(2) H(4) + H(2) O = Fe (OH)(2) + NH(3)
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- H(2) S + HNO(3) = NO + S + H(2) O
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- Balance the chemical equation by the oxidation number method. (ii) ...
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- Balance the following equation by oxidation number method K(2)Cr(2)O...
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- MnO(4)^(-) + C(2) O(4)^(2-) + H^(+) = CO(2) + Mn^(2+) + H(2) O
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- Cr(2) O(7)^(2-) + C(2) O(4)^(2-) + H^(+) = Cr^(3+) + CO(2) + H(2) O
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- Cr(OH)(3) + IO(3) + OH^(-) = I^(-) + CrO(4)^(2-) + H(2)O
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- KMnO(4) overset(Delta) to K(2)MnO(4)+MnO(2)+O(2)uarr
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- (NH(4))(2) Cr(2)O(7) = N(2) + H(2) O + Cr(2) O(3)
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