Oxidation-Reduction Dr. Ron Rusay Spring 2008

Oxidation-Reduction Oxidation is the loss of electrons. Reduction is the gain of electrons. The reactions occur together. One does not occur without the other. The terms are used relative to the change in the oxidation state or oxidation number of the reactant(s).

Oxidation Reduction Reactions

Oxidation State (Oxidation Number)

QUESTION

ANSWER 2) NO Section 4.9 Oxidation Reduction Reactions (p. 154) Oxygen almost always has an oxidation state of 2 when part of a compound. The exception is when it is part of a peroxide. For example, hydrogen peroxide H . Then it has an oxidation state of 2 - STUDENT CD: Understanding Concepts: Oxidation-Reduction Reactions – O 2 2 – 1 .

Zinc

QUESTION

ANSWER 3) 2 Section 4.9 Oxidation Reduction Reactio ns (p. 154) If an element is found on the reactant’s side, this is almost always a redox reaction, since an element usually becomes part of a compound during a chemical reaction. -

- 2 e- +2 e- Cu 2+  Cu (s) H2 (g)  2 H + Number of electrons gained must equal the number of electrons lost. - 2 e- +2 e- Use oxidation numbers to determine what is oxidized and what is reduced. +2 e- Cu 2+  Cu (s) - 2 e- H2 (g)  2 H + Refer to Balancing Oxidation-Reduction Reactions

QUESTION

ANSWER 2) the oxidizing agent. Section 4.9 Oxidation Reduction Reactions (p. 154) Metals lose electrons, so they are oxidized, making the other reactant an oxidizing agent. - HMCLASS PREP: Figure 4.19

Balancing Redox Equations in acidic solutions 1) Determine the oxidation numbers of atoms in both reactants and products. 2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”. 3) Balance the “redox” atoms and charges (electron gain and loss must equal!). 4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).

Balancing Redox Equations Fe+2(aq)+ Cr2O72-(aq) +H+(aq)----> Fe3+(aq) + Cr3+(aq) + H2O(l) ? Fe 2+(aq)+ Cr2O72-(aq) +H+(aq)----> Fe 3+(aq) + Cr 3+(aq) + H2O(l) x = ? Cr ; 2x+7(-2) = -2; x = +6

Fe 2+(aq) ---> Fe 3+(aq) Balancing Redox Equations Fe 2+(aq) ---> Fe 3+(aq) Cr2O72-(aq) + --> Cr 3+(aq) Cr = (6+) -e - 2 6 e - 6 (Fe 2+(aq) -e - ---> Fe3+(aq)) 6 Fe 2+(aq) ---> 6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)

Balancing Redox Equations 6 Fe2+(aq) ---> 6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq) 6 Fe2+(aq)+ Cr2O72-(aq) + ? 2nd H+(aq) ----> 6 Fe3+(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l) Oxygen = 7 2nd (Hydrogen) = 14

Balancing Redox Equations Completely Balanced Equation: 6 Fe2+(aq)+ Cr2O72-(aq) + 14 H+(aq) ----> 6 Fe3+(aq) + 2 Cr3+(aq)+ 7 H2O(l)

Balancing Redox Equations in basic solutions 1) Determine oxidation numbers of atoms in Reactants and Products 2) Identify and select out those which change oxidation number into separate “half reactions” 3) Balance redox atoms and charges (electron gain and loss must equal!) 4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water

Balancing Redox Equations in basic solutions MnO2 (aq)+ ClO31-(aq) + OH 1-aq) -> MnO41- (aq)+ Cl 1-(aq) + H2O(l) Mn4+ (MnO2) ---> Mn7+ (MnO4 ) 1- Cl+5 (ClO3 ) 1-+ 6 e- ---> Cl 1-

Balancing Redox Equations in basic solutions Electronically Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 6 e - ----> 2 MnO4 1- + Cl 1- + 6 e-

Balancing Redox Equations in basic solutions Completely Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 2 OH 1- (aq)----> 2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l) 9 O in product

MnO4– + C2O42–  MnO2 + CO32– (basic solution) QUESTION Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4– could be used to analyze samples for oxalate. MnO4– + C2O42–  MnO2 + CO32– (basic solution) When properly balanced, how many OH– are present? 1. 1 2. 2 3. 3 4. 4

ANSWER Choice 4 is correct. The redox equation could be balanced according to the steps for an acid solution, but then OH– must be added to neutralize the H+ ions. When done properly 4 OH– ions will be present in the basic equation. Section 4.10: Balancing Oxidation–Reduction Equations