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Chapter 19 Oxidation–Reduction (Redox) Reactions. Electron Transfer Reactions. The reaction between zinc and copper(II) ions is spontaneous. Electron Transfer Reactions. Zinc is said to be oxidized to zinc ion and copper(II) ions is said to be reduced to copper. Electron Transfer Reactions. - PowerPoint PPT Presentation
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www.cengage.com/chemistry/cracolice
Mark S. CracoliceEdward I. Peters
Mark S. Cracolice • The University of Montana
Chapter 19Oxidation–Reduction
(Redox) Reactions
Electron Transfer Reactions
The reaction between zinc and copper(II) ions is spontaneous.
Electron Transfer Reactions
Zinc is said to be oxidized to zinc ion and copper(II) ions is said to be reduced to copper.
Electron Transfer Reactions
The same reaction between zinc and copper(II) ions can be performed using the following arrangement.
Electron Transfer ReactionsWhen the two metallic rods are connected, zinc is oxidized to zinc
ions and releases electrons at the zinc electrode and copper (II) ions collect electrons from the copper electrode to become copper.
Electron Transfer Reactions
The overall oxidation-reduction reaction can be separated into two half reactions.
Zn (s) Zn2+ (aq) + 2e- ( at the zinc rod)
Cu2+ (aq) + 2e- Cu (s) ( at the copper rod)_________________________________________
Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)
Electron Transfer Reactions
The process of loosing electrons is called oxidation.
The chemical change that occurs at the zinc electrode is oxidation, zinc is oxidized to Zn2+.
Zn(s) Zn2+(aq) + 2 e–
The process of gaining electrons is called reduction.
Copper (II) cations are reduced to Cu at the copper rod.Cu2+(aq) + 2 e– Cu(s)
Electron Transfer ReactionsIf half-reaction equations are added algebraically, the result is
the net ionic equation for the oxidation–reduction reaction:
Zn(s) Zn2+(aq) + 2 e–
Cu2+(aq) + 2 e– Cu(s)
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
The chemical change is an electron transfer reaction; electrons have been transferred from zinc atoms to copper(II) ions.
Voltaic (Galvanic) Cells
Voltaic (Galvanic) CellA cell in which a spontaneous chemical reaction is used
to produce electricity.
Voltaic (Galvanic) CellsAnode: Electrode at which oxidation occurs. The species being
oxidized releases electrons into the electrode.Cathode: Electrode at which reduction occurs. The species
being reduced collect electrons from the electrode
Commercial Galvanic CellsAlkaline battery One electrode is zinc in contact with a gel of potassium
hydroxide.One electrode is graphite in contact with a mixture of manganese
(IV) oxide, graphite and potassium hydroxide.
Reaction at the anode:Zn (s) + 2 OH- (aq) ZnO (s) + H2O (l) + 2e
Reaction at the cathode:2 MnO2 (s) + H2O (l) + 2e Mn2O3 (s) + 2 OH-
Commercial Galvanic CellsLead battery One electrode is lead in contact with a sulfuric acid solution One electrode is lead in contact with lead (IV) oxide and sulfuric
acid
Reaction at the anode:Pb (s) + SO4
2- (aq) PbSO4 (s) + 2e
Reaction at the cathode:PbO2 (s) + 4 H+ + SO4
2- (aq) + 2e PbSO4 (s) + 2 H2O
Lead battery is rechargeable.
Electrolytic CellsElectrolytic Cell
In an electrolytic cell, current supplied by an external source is used to drive a non-spontaneous redox reaction
ComponentsAn electrolyte and two electrodes.
When the electrodes are connected to an outside source of electricity, they become charged, one positively and one
negatively. The ions in the electrolyte move to the oppositely charged electrode, where chemical reactions occur. The
movement of ions is an electric current in solution.
Electrolytic CellsElectrolysis of liquid sodium chloride. Two electrodes are connected to an external electric source and dipped in liquid sodium chloride.
Electrolytic CellsAt Negative Electrode: sodium ions Na+ are attracted to the negative electrode, accept electrons from the electrode and are reduced to sodium. The reaction is reduction so the electrode is called cathode.
Na+ + e- Na
At Positive Electrode: Chloride ions Cl- are attracted to the positive electrode, where they are oxidized to chlorine gas.
The reaction is oxidation, so the electrode is an anode.
2 Cl- + 2 e- Cl2
Oxidation Numbers & Redox Reactions
Electron-transfer reaction is called redox reaction.
The reactant that gains electrons is called the oxidizing agent.
The reactant that loses electrons is called the reducing agent.
In the reaction, the oxidizing agent is reduced, and the reducing agent oxidized.
To determine which species has gained and which has lost electrons, it is useful to determine the oxidation states, or
oxidation numbers of reactants and products.
Rules for Determination of Oxidation NumbersThe oxidation number of any elemental substance is 0The oxidation number of a monatomic ion is the same as the
charge on the ionThe oxidation number of combined oxygen is -2, except in
peroxide(-1), superoxide(-1/2), and OF2 (+2)The oxidation number of combined hydrogen is +1(except
hydride ion , H- ).In any molecular or ionic species the sum of the oxidation
numbers of all atoms in a molecule or polyatomic ion is equal to the charge on the species.
Oxidation is an increase in oxidation number.Reduction is a decrease in oxidation number.•
Oxidizing and Reducing AgentsOxidizing Agent (Oxidizer)
Species that removes electrons in a redox reaction(the species that is reduced).
Reducing Agent (Reducer)Species from which electrons are removed
(the species that is oxidized).
Strengths of Ox and Red AgentsStrong oxidizing agent
A species with a strong attraction for electrons.
Weak oxidizing agentA species that attracts electrons only slightly.
Strong reducing agentA species that releases electrons readily.
Weak reducing agentA species that does not easily give up its electrons.
Strengths of Oxidizing and Reducing Agents
Predicting Redox Reactions
Writing Redox EquationsWriting Redox Equations for
Half-Reactions in Acidic Solutions
1. After identifying the element oxidized or reduced, write a skeleton half-reaction equation with the element in its original form (element, monatomic ion, or part of a polyatomic ion or compound) on the left and its final form on the right.
2. Balance the element oxidized or reduced.
3. Balance elements other than hydrogen or oxygen, if any.
Writing Redox EquationsWriting Redox Equations for
Half-Reactions in Acidic Solutions
4. Balance oxygen by adding water molecules where necessary.
5. Balance hydrogen by adding H+ where necessary.
6. Balance charge by adding electrons to the more positive side.
7. Recheck the equation to be sure it is balanced in both number of atoms and charge.
Writing Redox EquationsExample:Iron(II) ion solution and a permanganate ion (MnO4
–) solution are combined in an acidic solution. Manganese(II) ion and iron(III) ion are produced. Write a balanced net ionic equation for the reaction.
Solution:First, we summarize the reaction description:Reactants: Fe2+(aq) and MnO4
–(aq)The reaction is in acidic solution, so H+(aq) is also a reactantProducts: Fe3+(aq) and Mn2+(aq)
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 1: Write skeleton half-reaction equations
Fe2+(aq) Fe3+(aq)MnO4
–(aq) Mn2+(aq)
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 2: Balance element oxidized/reduced
Fe2+(aq) Fe3+(aq) MnO4–(aq) Mn2+(aq)
Both Fe and Mn are already balanced
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 3: Balance elements other than H, O
Fe2+(aq) Fe3+(aq) MnO4–(aq) Mn2+(aq)
There are none
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 4: Balance O by adding H2O
Fe2+(aq) Fe3+(aq) MnO4–(aq) Mn2+(aq) + 4 H2O(l)
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 5: Balance H by adding H+
Fe2+(aq) Fe3+(aq) 8 H+ + MnO4–(aq) Mn2+(aq) + 4 H2O(l)
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 6: Balance charge by adding electrons
Fe2+(aq) Fe3+(aq) + e–
5 e– + 8 H+ + MnO4–(aq) Mn2+(aq) + 4 H2O(l)
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Step 7: Check for charge and atom balance
Fe2+(aq) Fe3+(aq) + e–
2+ each side, 1 Fe each side5 e– + 8 H+ + MnO4
–(aq) Mn2+(aq) + 4 H2O(l)2+ each side, 8 H, 1 Mn, 4 O each side
Writing Redox EquationsReactants: Fe2+(aq), MnO4
–(aq), H+(aq) Products: Fe3+(aq), Mn2+(aq)
Solution:Add the half reactions to yield the complete net ionic equation
5 × [Fe2+(aq) Fe3+(aq) + e–]5 e– + 8 H+ + MnO4
–(aq) Mn2+(aq) + 4 H2O(l)
5 Fe2+(aq) + 8 H+ + MnO4–(aq) 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
Homework• Homework: 3, 9, 11, 17, 19, 21, 23,25, 29, 49, 59• Write net equations for all reactions of the Experiment 13
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