CHAPTER 15 Electro- chemistry 15.3 Balancing Redox Equations

2 Redox reactions Zn(s) CuSO 4 (aq) Cu(s) deposit Zn(s) + CuSO 4 → ZnSO 4 (aq) + Cu(s)

Balancing Redox Equations Redox reactions Zn(s) + CuSO 4 → ZnSO 4 (aq) + Cu(s) We saw that this is a redox reaction in which: some elements lose electrons; they are oxidized other elements gain electrons; they are reduced We learned how to determine oxidation numbers

Balancing Redox Equations Redox reactions Zn(s) + CuSO 4 → ZnSO 4 (aq) + Cu(s) We saw that this is a redox reaction in which: some elements lose electrons; they are oxidized other elements gain electrons; they are reduced We learned how to determine oxidation numbers Now we look at how to balance redox reactions

Balancing Redox Equations Two methods At the end, both mass and charge have to be balanced There are two methods: 1) The oxidation number method 2) The half-reaction method

Balancing Redox Equations Step 1 Assign oxidation numbers for all atoms Step 2 Identify the atoms that are oxidized, and atoms that are reduced Step 3 Adjust coefficients for atoms whose oxidation numbers change, then make sure that the rule above is observed Step 4 Check the overall mass balance The oxidation number method Increase in oxidation number for oxidized atoms Decrease in oxidation number for reduced atoms =

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l)

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 1Assign oxidation numbers for all atoms H N O 3 + Cu 2 O → Cu (N O 3 ) 2 + N O + H 2 O –2 +1 – –2 +2 –2 +1 –2

Balancing Redox Equations oxidation reduction Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 2Identify the atoms that are oxidized, and atoms that are reduced H N O 3 + Cu 2 O → Cu (N O 3 ) 2 + N O + H 2 O Oxidation numbers do not change –2 +1 – –2 +2 –2 +1 –2 N is reducedits oxidation number goes from +5 to +2 Cu is oxidized its oxidation number goes from +1 to +2

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 3Balance all atoms whose oxidation numbers have changed H N O 3 + Cu 2 O → 2Cu (N O 3 ) 2 + N O + H 2 O oxidation reduction –2 +1 – –2 +2 –2 +1 –2 N is already balanced: 1 atom on each side Cu needs to be adjusted with a coefficient of 2 N is reducedits oxidation number goes from +5 to +2 Cu is oxidized its oxidation number goes from +1 to +2

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 3Balance all atoms whose oxidation numbers have changed Balance the number of electrons (using oxidation numbers) H N O 3 + Cu 2 O → 2Cu (N O 3 ) 2 + N O + H 2 O oxidation reduction N is reducedits oxidation number goes from +5 to +2; it gains 3 electrons Cu is oxidized its oxidation number goes from +1 to +2; it loses 2 electrons because of the coefficient –2 +1 – –2 +2 –2 +1 –2

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 3Balance all atoms whose oxidation numbers have changed Balance the number of electrons (using oxidation numbers) H N O 3 + Cu 2 O → 2Cu (N O 3 ) 2 + N O + H 2 O –2 e – + 3 e – N is reducedits oxidation number goes from +5 to +2; it gains 3 electrons Cu is oxidized its oxidation number goes from +1 to +2; it loses 2 electrons The number of electrons transferred must be the same –2 +1 – –2 +2 –2 +1 –2

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 3Balance all atoms whose oxidation numbers have changed Balance the number of electrons (using oxidation numbers) 2H N O 3 + 3Cu 2 O → 6Cu (N O 3 ) 2 + 2N O + H 2 O N is reducedits oxidation number goes from +5 to +2; it gains 3 electrons Cu is oxidized its oxidation number goes from +1 to +2; it loses 2 electrons The number of electrons transferred must be the same (+ 3 e – ) x 2 = +6 e – –2 +1 – –2 +2 (–2 e – ) x 3 = –6 e –

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 4Check the overall mass balance 12HNO 3 + 3Cu 2 O → 6Cu(NO 3 ) 2 + 2NO + 7H 2 O H2 N2 Cu6 O7 H 2 N14 Cu 6 O39 Masses are not balanced yet Start balancing N

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 4Check the overall mass balance 14HNO 3 + 3Cu 2 O → 6Cu(NO 3 ) 2 + 2NO + 7H 2 O H14 N14 Cu 6 O45 H 2 N14 Cu 6 O39 Masses are not balanced yet Now balance O and H

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step 4Check the overall mass balance 14HNO 3 + 3Cu 2 O → 6Cu(NO 3 ) 2 + 2NO + 7H 2 O H14 N14 Cu 6 O45 H14 N14 Cu 6 O45 Masses are now balanced

Balancing Redox Equations Using the oxidation number method, balance the equation: HNO 3 (aq) + Cu 2 O(s) → Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) 14HNO 3 + 3Cu 2 O → 6Cu(NO 3 ) 2 + 2NO + 7H 2 O Answer:

Balancing Redox Equations Two methods At the end, both mass and charge have to be balanced There are two methods: 1) The oxidation number method 2) The half-reaction method The redox reaction is split into two half-reactions: the oxidation reaction, and the reduction reaction.

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + Step 5Balance the charge for both half-reactions by adding electrons The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + Step 5Balance the charge for both half-reactions by adding electrons Step 6Adjust coefficients to balance the number of electrons transferred The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + Step 5Balance the charge for both half-reactions by adding electrons Step 6Adjust coefficients to balance the number of electrons transferred Step 7Combine the two half-reactions, and add the spectator ions The half-reaction method

Balancing Redox Equations Step 1Write the unbalanced equation showing explicitly all ions Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) Step 3Write down the two unbalanced half-reactions Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + Step 5Balance the charge for both half-reactions by adding electrons Step 6Adjust coefficients to balance the number of electrons transferred Step 7Combine the two half-reactions, and add the spectator ions Step 8Simplify and check that both mass and charge are balanced The half-reaction method

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 1Write the unbalanced equation showing explicitly all ions Zn(s) + Cu 2+ (aq) + SO 4 –2 (aq) → Zn 2+ (aq) + SO 4 –2 (aq) + Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 2Identify the atoms that are oxidized, and atoms that are reduced Find spectator ions (unchanged oxidation numbers) 0 +2 –2 +2 –2 0 oxidation reduction spectator Zn(s) + Cu 2+ (aq) + SO 4 –2 (aq) → Zn 2+ (aq) + SO 4 –2 (aq) + Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 3Write down the unbalanced half-reactions 0 +2 –2 +2 –2 0 oxidation reduction Zn(s) + Cu 2+ (aq) + SO 4 –2 (aq) → Zn 2+ (aq) + SO 4 –2 (aq) + Cu(s) Oxidation:Reduction: Zn(s) → Zn 2+ (aq)Cu 2+ (aq) → Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 4Balance mass with elements other than oxygen and hydrogen Balance oxygen by adding H 2 O, then balance hydrogen with H + Zn(s) → Zn 2+ (aq) mass already balanced Oxidation:Reduction: Cu 2+ (aq) → Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 5Balance the charge for both half-reactions by adding electrons Zn(s) → Zn 2+ (aq) + 2e – Oxidation:Reduction: Cu 2+ (aq) + 2e – → Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 6 Adjust coefficients to balance the number of electrons transferred Zn(s) → Zn 2+ (aq) + 2e – already the same in both half-reactions Oxidation:Reduction: Cu 2+ (aq) + 2e – → Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 7 Combine the two half-reactions, and add the spectator ions Zn(s) → Zn 2+ (aq) + 2e – Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Oxidation:Reduction: Cu 2+ (aq) + 2e – → Cu(s)

Balancing Redox Equations Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Step 8 Simplify and check that both mass and charge are balanced Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Using the half-reaction method, we were able to determine that 2 electrons were transferred The full equation could not provide us with that information

Balancing Redox Equations At the end, both mass and charge have to be balanced There are two methods: 1) The oxidation number method 2) The half-reaction method