In the following reaction what is the change of oxidation state of $P b$ ? $Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)$

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Solution

$Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)$
0 +2 0 +2
Zinc metal is oxidised to ${Z n}^{2 +}$
$Z n$ is the reducer.

${P b}^{2 +}$ is the oxidiser.
${P b}^{2 +}$ is reduced to $P b$

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Similar questions

Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

Z n (s) ⟶ {Z n}^{2 +} (a q) + 2 e^{-}

{P b}^{2 +} (a q) + 2 e^{-} ⟶ P b (s)

M(s) +

PbCl

_{2} (a q)

\to

Pb(s) +

MCl

_{2} (a q)

S O_{2}

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