consider this unbalanced reaction.\nce^{3 +}(aq)+pbo_{2}(s)+so_{4}^{2 -}(aq)\rightarrow pbso_{4}(s)+ce^{4…

Explanation:

Step1: Identify oxidation - reduction half - reactions

Oxidation: $\ce{Ce^{3+}(aq)\to Ce^{4+}(aq)}$ Reduction: $\ce{PbO2(s) + SO4^{2 - }(aq)\to PbSO4(s)}$

Step2: Balance oxidation half - reaction

$\ce{Ce^{3+}(aq)\to Ce^{4+}(aq)+e^{-}}$

Step3: Balance reduction half - reaction

First, balance oxygen by adding $\ce{H2O}$: $\ce{PbO2(s) + SO4^{2 - }(aq)\to PbSO4(s)+2H2O(l)}$ Then balance hydrogen by adding $\ce{H^{+}}$: $\ce{PbO2(s) + SO4^{2 - }(aq)+4H^{+}(aq)\to PbSO4(s)+2H2O(l)}$ Finally, balance charge by adding electrons: $\ce{PbO2(s) + SO4^{2 - }(aq)+4H^{+}(aq)+2e^{-}\to PbSO4(s)+2H2O(l)}$

Step4: Multiply half - reactions to equalize electrons

Multiply oxidation half - reaction by 2: $2\ce{Ce^{3+}(aq)\to 2Ce^{4+}(aq)+2e^{-}}$

Step5: Combine half - reactions

$2\ce{Ce^{3+}(aq)}+\ce{PbO2(s)}+\ce{SO4^{2 - }(aq)} + 4\ce{H^{+}(aq)}\to\ce{PbSO4(s)}+2\ce{Ce^{4+}(aq)}+2\ce{H2O(l)}$

Answer:

$2\ce{Ce^{3+}(aq)}+\ce{PbO2(s)}+\ce{SO4^{2 - }(aq)} + 4\ce{H^{+}(aq)}\to\ce{PbSO4(s)}+2\ce{Ce^{4+}(aq)}+2\ce{H2O(l)}$