Balancing Oxidation-Reduction Reactions Dr. Ron Rusay Balancing Oxidation-Reduction Reactions

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).

Aqueous Reactions: Oxidation - Reduction In the following reaction, identify what is being oxidized and what is being reduced. What is the total number of electrons involved in the process?

Oxidation Reduction Reactions

QUESTION In a redox reaction, oxidation and reduction must both occur. Which statement provides an accurate premise of redox chemistry? The substance that is oxidized must be the oxidizing agent. The substance that is oxidized must gain electrons. The substance that is oxidized must have a higher oxidation number afterwards. The substance that is oxidized must combine with oxygen.

ANSWER In a redox reaction, oxidation and reduction must both occur. Which statement provides an accurate premise of redox chemistry? The substance that is oxidized must be the oxidizing agent. The substance that is oxidized must gain electrons. The substance that is oxidized must have a higher oxidation number afterwards. The substance that is oxidized must combine with oxygen.

QUESTION

ANSWER B) NO 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 .

What is the oxidation number of chromium in ammonium dichromate? QUESTION What is the oxidation number of chromium in ammonium dichromate? A) +3 B) +4 C) +5 D) +6

What is the oxidation number of chromium in ammonium dichromate? ANSWER What is the oxidation number of chromium in ammonium dichromate? A) +3 B) +4 C) +5 D) +6 Correct Answer: D Question Number: 1 (NH4)2Cr2O7

Zinc

Reactivity Tables (usually reducing) show relative reactivities: In the examples from the previous slide, the acid solution (H+) will react with anything below it in the Table but not above. Nickel and Zinc, ….but not Copper.

A) I and II B) II and III C) I and IV D) III and IV QUESTION Select all redox reactions by looking for a change in oxidation number as reactants are converted to products. I) Ca + 2 H2O → Ca(OH)2 + H2 II) CaO + H2O → Ca(OH)2 III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O IV) Cl2 + 2 KBr → Br2 + 2 KCl A) I and II B) II and III C) I and IV D) III and IV

A) I and II B) II and III C) I and IV D) III and IV ANSWER Select all redox reactions by looking for a change in oxidation number as reactants are converted to products. I) Ca + 2 H2O → Ca(OH)2 + H2 II) CaO + H2O → Ca(OH)2 III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O IV) Cl2 + 2 KBr → Br2 + 2 KCl A) I and II B) II and III C) I and IV D) III and IV Correct Answer: C Question Number: 2

QUESTION

ANSWER C) 2 If an element is a reactant, this usually a redox reaction, since an element usually becomes part of a compound during a chemical reaction. Cu + 2AgNO3  2Ag + Cu(NO3)2 N2 + 3H2  2NH3

- 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 B) the oxidizing agent. 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).

Fe 3+(aq) + Cr 3+(aq) + H2O(l) 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) ? Cr oxidation number? x = ? Cr ; 2x+7(-2) = -2; x = +6

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)

Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) QUESTION Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation: Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16  

Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) ANSWER Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation: Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16   Correct Answer: D Question Number: 3

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-

2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l) 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 2 3 4

ANSWER D). 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.