Cl(2) + KOH = KOCl + KCl + H(2)O...

`Cl_(2) + KOH = KOCl + KCl + H_(2)O`

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To balance the redox reaction \( \text{Cl}_2 + \text{KOH} \rightarrow \text{KOCl} + \text{KCl} + \text{H}_2\text{O} \), we will follow these steps: ### Step 1: Identify Oxidation States - In \( \text{Cl}_2 \), chlorine has an oxidation state of 0. - In \( \text{KOCl} \), the oxidation state of Cl is +1. - In \( \text{KCl} \), the oxidation state of Cl is -1. This indicates that chlorine is undergoing a disproportionation reaction, where it is both oxidized and reduced. ...

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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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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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MnO(4)^(2-) + H^(+) = MnO(4)^(-) + MnO(2) + H(2) O
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