How is $cuprous ion$ $"cuprous ion"$ oxidized by bleach to give $cupric ion$ $"cupric ion"$ ?

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Answer 1 · 2016-09-11T07:48:38

This is a redox reaction; cuprous ion is oxidized by bleach to cupric ion.

$C l O^{-} + 2 H^{+} + 2 e^{-} \to C l^{-} + H_{2} O (l)$ $ClO^(-) +2H^(+) + 2e^(-) rarr Cl^(-) + H_2O(l)$ , $reduction$ $"reduction"$

$C u^{+} \to C u^{2 +} + 1 e^{-}$ $Cu^(+) rarr Cu^(2+) + 1e^(-)$ , $oxidation$ $"oxidation"$

$2 C u^{+} + C l O^{-} + 2 H^{+} \to 2 C u^{2 +} + C l^{-} + H_{2} O (l)$ $2Cu^(+) + ClO^(-) +2H^(+)rarr 2Cu^(2+) + Cl^(-) + H_2O(l)$ .

Charge and mass are balanced, so this equation is reasonable.

How is cuprous ion"cuprous ion" oxidized by bleach to give cupric ion"cupric ion"?