(a) We have the following redox reaction : H 2 O 2 ( a q ) + C l O 2 ( a q ) → C l O 2 − ( a q ) + O 2 ( g ) Step 1: Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. O x i d a t i o n : H 2 O 2 ( a q ) → O 2 ( g ) R e d u c t i o n : C l O 2 ( a q ) → C l O 2 − ( a q ) Step 2: Balance the elements, excluding H, O. O x i d a t i o n : H 2 O 2 ( a q ) → O 2 ( g ) R e d u c t i o n : C l O 2 ( a q ) → C l O 2 − ( a q ) Step 3: Balance the elements H, O. O x i d a t i o n : H 2 O 2 ( a q ) + 2 H O − ( a q ) → O 2 ( g ) + 2 H 2 O ( l ) R e d u c t i o n : C l O 2 ( a q ) → C l O 2 − ( a q ) In these half-reactions, all elements are balanced, including H, O. Hence, in the next step we need to balance the charge. Step 4: Balance each half-reaction with respect to charge by adding electrons. O x i d a t i o n : H 2 O 2 ( a q ) + 2 H O − ( a q ) → O 2 ( g ) + 2 H 2 O ( l ) + 2 e − R e d u c t i o n : C l O 2 ( a q ) + 1 e − → C l O 2 − ( a q ) Step 5: Make the number of electrons in both half-reactions equal by multiplying one or both half reactions by a small whole number. O x i d a t i o n : [ H 2 O 2 ( a q ) + 2 H O − ( a q ) → O 2 ( g ) + 2 H 2 O ( l ) + 2 e − ] × 1 R e d u c t i o n : [ C l O 2 ( a q ) + 1 e − → C l O 2 − ( a q ) ] × 2 Add these two reactions to get the balanced reaction. H 2 O 2 ( a q ) + 2 H O − ( a q ) + 2 C l O 2 ( a q ) → O 2 ( g ) + 2 H 2 O ( l ) + 2 C l O 2 − ( a q ) (b) We have the following redox reaction: A l ( s ) + M n O 4 − ( a q ) → M n O 2 ( s ) + A l ( O H ) 4 − ( a q ) Step 1: Separate the overall reaction into two half-reactions: one for the oxidation and one for the reduction. O x i d a t i o n : A l ( s ) → A l ( O H ) 4 − ( a q ) R e d u c t i o n : M n O 4 − ( a q ) → M n O 2 ( s ) Step 2: Balance the elements, excluding H, O. O x i d a t i o n : A l ( s ) → A l ( O H ) 4 − ( a q ) R e d u c t i o n : M n O 4 − ( a q ) → M n O 2 ( s ) Step 3: Balance the elements H, O

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