Equation is as follow,
SnCl₂ + PbCl₄ → SnCl₄ + PbCl₂
Oxidation State of Sn on left hand side in SnCl₂ is +2, while that in SnCl₄ on right hand side is +4. Means Sn has lost two electrons. Hence, it is oxidized and has worked as a reducing agent (has reduced Pb from Pb⁺⁴ to Pb⁺²).
Oxidation State of Pb on left hand side in PbCl₄ is +4, while that in PbCl₂ on right hand side is +2. Means Pb has gained two electrons. Hence, it is reduced and has worked as an oxidizing agent (has oxidized Sn from Sn⁺² to Sn⁺⁴).
