Can you represent the oxidation of sulfite ion, $S O_{3}^{2 -}$ $SO_3^(2-)$ , to give sulfate ion, $S O_{4}^{2 -}$ $SO_4^(2-)$ , accompanied by the reduction of $C u^{2 +}$ $Cu^(2+)$ to copper metal?

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Answer 1 · 2016-10-18T03:21:47

$S O_{3}^{2 -} + C u^{2 +} + 2 H O^{-} \to S O_{4}^{2 -} + C u (s) + H_{2} O (l)$ $SO_3^(2-) + Cu^(2+)+ 2HO^(-) rarr SO_4^(2-) +Cu(s) + H_2O(l)$

When basic conditions are specified, it is normally best to balance the equations in acid, and then modify the final equations.

$Sulfur (IV+)$ $"Sulfur (IV+)"$ $\to$ $rarr$ $Sulfur(VI+) :$ $"Sulfur(VI+)":$

$S O_{3}^{2 -} + H_{2} O \to S O_{4}^{2 -} + 2 H^{+} + 2 e^{-}$ $SO_3^(2-) + H_2Orarr SO_4^(2-) +2H^(+) + 2e^(-)$

For basic conditions, add $2 \times H O^{-}$ $2xxHO^-$ to EACH side $(H O^{-} + H^{+} \to H_{2} O)$ $(HO^(-) + H^(+)rarrH_2O)$ :

$S O_{3}^{2 -} + 2 H O^{-} \to S O_{4}^{2 -} + H_{2} O + 2 e^{-}$ $SO_3^(2-) + 2HO^(-) rarr SO_4^(2-) +H_2O + 2e^(-)$

$Copper (II+)$ $"Copper (II+)"$ $\to$ $rarr$ $Copper (0) :$ $"Copper (0)":$

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

You have asked another such question recently. Do you follow the method of doing redox equations?

Can you represent the oxidation of sulfite ion, SO2−3SO_3^(2-), to give sulfate ion, SO2−4SO_4^(2-), accompanied by the reduction of Cu2+Cu^(2+) to copper metal?