Chapter 21 Lecture 2

# Chapter 21 Lecture 2 - Example Balance the skeleton redox...

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Example: Balance the skeleton redox equation in acidic solution: 3+ Ce 4+ VO + 3+ V + Ce 2 + Ce 1. Identify redox couples: VO 2 + /V 3+ and Ce 4+ /Ce 3+ Half-reactions: V 3+ VO 2 + and Ce 4+ Ce 3+ 2. Balance the half reactions: V 3+ VO 2 + (V is balanced) V 3+ + 2H 2 O VO 2 + (balance O) + V 3+ + 2H 2 O VO 2 + + 4H + (balance H) V 3+ + 2H 2 O VO 2 + + 4H + + 2e - (balance charge) e 4+ Ce 3+ e balanced) Ce (Ce is balanced) Ce 4+ + 1e - Ce 3+ (balance charge) 3. Multiply the 2 nd half-reaction by 2 to get 2e- 2 Ce 4+ + 2 e - 2 Ce 3+ 4. Add the half-reactions + 2H VO 4H 2e V 3+ + 2H 2 O 2 + + 4H + + 2e - +2 C e 4+ + 2e - 2Ce 3+ 3+ + 2H + 2Ce 4+ + 2e - VO + + 4H + + 2e - +2Ce 3+ 21-10 V 2H 2 O 2Ce 2e 2 4H 2e 2Ce V 3+ + 2Ce 4+ + 2H 2 O VO 2 + +2Ce 3+ + 4H +

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Balancing Redox Reactions in Basic Solutions by the Half-Reaction Method 1–4. The same four steps as those in acid solutions 5. Add OH - on both sides of the equation in order to neutralize the H + , and cancel the water molecules if necessary PROBLEM: Permanganate ion reacts in basic solution with the oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO 4 and Na 2 C 2 O 4 in basic solution: MnO 4 - ( aq ) + C 2 O 4 2- ( aq ) MnO 2 ( s ) + CO 3 2- ( aq ) PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: MnO 4 - MnO 2 C 2 O 4 2- CO 3 2- 7 4 MnO 4 - MnO 2 C 2 O 4 2- CO 3 2- 2 MnO - MnO + 2H O 4H + + 2- O 2- 2H 4H + +7 +4 +3 +4 21-11 4 2 2 C 2 O 4 2 CO 3 + 2H 2 O + 4H +3e - +2e -
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Chapter 21 Lecture 2 - Example Balance the skeleton redox...

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