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Electrochemical Reactions Redox reaction: electrons transferred from one species to another Oxidation ≡ loss of electrons Reduction ≡ gain of electrons
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What is reduced is the oxidizing agent H + oxidizes Zn by taking electrons from it What is oxidized is the reducing agent Zn reduces H + by giving it electrons Oxidation and Reduction
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Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn is oxidized; Zn is reducing agentZn Zn 2+ + 2e - Cu 2+ is reduced; Cu 2+ is oxidizing agent Cu 2+ + 2e - Cu Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu 2+ + 2e - Ag + + 1e - AgAg + is reduced Ag + is oxidizing agent
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Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.
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Assigning Oxidation Numbers 1.Elements in their elemental form have an oxidation number of 0. 2.The oxidation number of a monatomic ion is the same as its charge. 3.Nonmetals tend to have negative oxidation numbers. 4.The sum of the oxidation numbers in a neutral compound is 0. 5.The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
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Balancing Redox Equations Neutral solution: Cr (s) + Ag + (aq) ⇌ Cr 3+ (aq) + Ag (s)
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Balancing Redox Equations Acidic solution: MnO 4 − (aq) + C 2 O 4 2− (aq) Mn 2+ (aq) + CO 2 (aq)
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Electrochemistry The Half-Reaction Method Consider the reaction between MnO 4 − and C 2 O 4 2− : MnO 4 − (aq) + C 2 O 4 2− (aq) Mn 2+ (aq) + CO 2 (aq) Balancing Redox Equations Fig 20.2
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Electrochemistry The Half-Reaction Method First, we assign oxidation numbers: MnO 4 − + C 2 O 4 2- Mn 2+ + CO 2 +7+3+4+2 Manganese goes from +7 to +2, it is reduced Carbon goes from +3 to +4, it is oxidized
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Electrochemistry Oxidation Half-Reaction C 2 O 4 2− CO 2 To balance the carbon, we add a coefficient of 2: C 2 O 4 2− 2 CO 2
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Electrochemistry Oxidation Half-Reaction C 2 O 4 2− 2 CO 2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C 2 O 4 2− 2 CO 2 + 2 e −
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Electrochemistry Reduction Half-Reaction MnO 4 − Mn 2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO 4 − Mn 2+ + 4 H 2 O
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Electrochemistry © 2009, Prentice- Hall, Inc. Reduction Half-Reaction MnO 4 − Mn 2+ + 4 H 2 O To balance the hydrogen, we add 8 H + to the left side. 8 H + + MnO 4 − Mn 2+ + 4 H 2 O
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Electrochemistry Reduction Half-Reaction 8 H + + MnO 4 − Mn 2+ + 4 H 2 O To balance the charge, we add 5 e − to the left side. 5 e − + 8 H + + MnO 4 − Mn 2+ + 4 H 2 O
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Electrochemistry Combining the Half-Reactions Now we evaluate the two half-reactions together: C 2 O 4 2− 2 CO 2 + 2 e − 5 e − + 8 H + + MnO 4 − Mn 2+ + 4 H 2 O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
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Electrochemistry Combining the Half-Reactions 5 C 2 O 4 2− 10 CO 2 + 10 e − 10 e − + 16 H + + 2 MnO 4 − 2 Mn 2+ + 8 H 2 O When we add these together, we get: 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− 2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e −
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Electrochemistry Combining the Half-Reactions 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− 2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− 2 Mn 2+ + 8 H 2 O + 10 CO 2
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Balancing Redox Equations Basic solution: V 2+ (aq) + V(OH) 4 + (aq) ⇌ VO 2+ (aq)
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