Redox Reactions Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004 Lecture #12

This lecture covers two topics: Assignment of oxidation numbers Balancing of redox equations

Oxidation - A process by which a substance (reductant) gives up electrons to another substance (oxidizing agent). The reductant is oxidized. Reduction - A process by which a substance (oxidant) accept electrons from another substance (reducing agent). In a chemical reaction, the total number of electrons are conserved; so also are the number of charges. Thus, it is convenient to assign fictitious charges to the atoms in a molecule and call them oxidation numbers. Oxidation numbers are chosen so that (a) conservation laws are obeyed, and (b) in ionic compounds the sum of oxidation numbers on the atoms coincides with the charge on the ion.

Rules for Assigning Oxidation Numbers: 1.The oxidation numbers of the atoms in a neutral molecule must add up to zero, and those in an ion must add up to the charge on the ion. 2.Alkali metal atoms have oxidation number +1, alkaline earth atoms +2, in their compounds. 3.Fluorine always has oxidation number -1 in its compounds. The other halogens have oxidation number -1 in their compounds except those with oxygen and with other halogens, where they can have positive oxidation number.

Oxidation Numbers (cont.): 4.Hydrogen is assigned oxidation number +1 in its compounds except in metal hydrides such as LiH, where convention (2) takes precedence and hydrogen has oxidation number Oxygen is assigned oxidation number -2 in compounds. There are two exceptions: in compounds with fluorine, convention 3 takes precedence, and in compounds that contain O - O bonds, conventions 2 and 4 take precedence. Thus the oxidation number of oxygen in OF 2 is +2: in peroxides (such as H 2 O 2 and Na 2 O 2 ) it is -1. In superoxides (such as KO 2 ) oxygen's oxidation number is -1/2.

Examples: NH 4 NO 3 N = H = N = O = B 2 H 6 B = H =

Examples (cont.): BaH 2 Ba = H = (S 4 O 6 ) 2- S = O =

Problem 12-1: Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (ox. no.) of each element in the following compounds. a) iron(III)chloride b) nitrogen dioxide c) sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the oxidation no. values in a neutral compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl 3 This compound is composed of monoatomic ions. The ox. no. of Cl - is -1, for a total of -3. Therefore the Fe must be +3. b) NO 2 c) H 2 SO 4

Problem 12-2:Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Rx: a) Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) b) S 8 (s) + 12 O 2 (g) 8 SO 3 (g) c) NiO(s) + CO(g) Ni(s) + CO 2 (g) Plan: First we assign an oxidation number (ox. no.) to each atom (or ion) based on the rules in Table 4.3. A reactant is the reducing agent if it contains an atom that is oxidized (ox. no. increased in the reaction). A reactant is the oxidizing agent if it contains an atom that is reduced ( ox. no. decreased). Solution: a) Assigning oxidation numbers: Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) HCl is the oxidizing agent, and Zn is the reducing agent!

Problem 12-2:Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S 8 (s) + 12 O 2 (g) 8 SO 3 (g) ____ is the reducing agent and ____ is the oxidizing agent. c) Assigning oxidation numbers: NiO (s) + CO (g) Ni (s) + CO 2 (g) ___ is the reducing agent and ____ is the oxidizing agent.

To proceed let us illustrate with the following problem: The objective is to find a set of 4 integer coefficients {x, y, z, w} which are as small as possible and also solve the set of atom conservation equations. These later state that no mass is created or destroyed and that the identity of the atoms is preserved. Thus for the above equation we have three atom balance equations, one each for N, H, and C. Problem 12-3: Balancing Redox Equations - I x NH 3 (g)+ y CH 4 (g) -> z HCN(g) + w H 2 (g)

Problem 12-3: These equations are: N: H: C: We see that we have 3 equations in 4 unknowns. Clearly we have one more unknown than equations. Thus, there are many possible valid solutions to the problem as stated. Mathematically, we say that the problem is underdetermined of ill-posed. We solve this problem by requiring that the set of values of the solution, {x,y,z,w} be the minimum set of integers. Now we have an additional equation which in most cases leads to an unique solution. Unfortunately, such a set of equations cannot be solved in one step. However for simple equations we give a straight forward manual procedure.

Problem 12-3: Two Step Hand Solution Here, we start with a guess for one of the coefficients, e.g. let x = 1, and then by back substitution we solve for all of the other coefficients. Hopefully, this will lead directly to the sought for solution in terms of the set of integers of minimum value that satisfy all of the atom balance equations. Often, our guess is not correct, and we are left with solutions that are rational numbers instead of integers. Now all we need to do is find the smallest common multiplier which will convert the solution to one of the set of integers of minimum values.

Let x = 1, Thus our solution is We verify it by substituting the numerical values of x, y, z and w into the atom balance equations. Problem 12-3: The Solution

Balancing Chemical Equations Involving Charged Species In this section we will show that balancing a chemical equation involving charged species closely resembles the previous approach and merely involves one additional equation to ensure that charge is balanced in addition to mass. One of the major difficulties of redox reactions is that it is often necessary to add H 2 O and H 3 O + to opposite sides of the reaction to balance the equation. As beginners we will assume that all pertinent species are provided and only coefficients are to be determined.

Example: Problem 12-4 t CuS + u NO v H 3 O + -> w Cu 2+ + x SO y NO + z H 2 O The objective is to find a set of integer coefficients {t, u, v, w, x, y, z} which are as small as possible and also solve the set of atom conservation equations.

For the above equation we have five atom balance equations, one each for Cu, S, N, O, H. These equations are: Cu: S: N: O: H: We see that we thus far have 5 equations in 7 unknowns, or since we have the three equalities t =w, t = x and u = y, we actually have 2 equations in 4 unknowns. Clearly we have two more unknowns than equations. Mathematically, we say that the problem is underdetermined.

In the case of charged species, the charges must also balance. Here is the equation for charge balance (CB): CB: Now that we have 6 equations in 7 unknowns, we can add the constraint that the solution set {x,y,z,t,u,v,w} be the minimum set of integers.

Problem 12-4: Hand Solution - Part I t = 1 = = This leaves 4 equations: Eliminate u Eliminate v Eliminate y Thus z = 4.

Problem 12-4: Hand Solution - Part II Now go backwards through the solutions solving for each variable. From z = 4 and y =2z/3 From v = y, From u = y, From v = u,

Problem 12-4: Hand Solution - Part III The above are a set of integers and rational numbers. By multiplying by 3 we can convert to the minimum set of integers. Thus the solution is _CuS + _NO _H 3 O+ -> _Cu 2+ + _(SO 4 ) 2- +_NO + _H 2 O

Problem 12-5: Redox Titration- I Volume (L) of KMnO 4 Solution Moles of KMnO 4 M (mol/L) Molar ratio in redox rxn. Chemical Formulas Moles of CaC 2 O 4 Problem: Calcium Oxalate was precipitated from 1.00 mL blood by the addition of Sodium Oxalate so the Ca 2+ conc. in the blood could be determined. This precipitate was dissolved in a sulfuric acid solution, which then required 2.05 mL of 4.88 x M KMnO 4 to reach the endpoint via the rxn. of Fig a) Calculate the moles of Ca 2+. b) Calculate the Ca 2+ conc. in blood. Plan: a) Calculate the moles of Ca 2+ in the H 2 SO 4 solution (and blood sample). b) Convert the Ca 2+ conc.into units of mg Ca 2+ / 100 mL blood. Moles of Ca +2 a) b) c)

Equation: 2 KMnO 4 (aq) + 5 CaC 2 O 4 (aq) + 8 H 2 SO 4 (aq) 2 MnSO 4 (aq) + K 2 SO 4 (aq) + 5 CaSO 4 (aq) + 10 CO 2 (g) + 8 H 2 O(l) Problem 12-5: Redox Titration - Calculation - II a) Moles of KMnO 4 b) Moles of CaC 2 O 4 c) Moles of Ca +2

Problem 12-5: Redox Titration - III Moles of Ca 2+ / 1 mL of blood Moles of Ca 2+ / 100 mL blood Mass (g) of Ca 2+ / 100 mL blood Mass (mg) of Ca 2+ / 100 mL blood multiply by 100 a) Calc of mol Ca 2+ per 100 mL M (g/mol) b) Calc of mass of Ca 2+ per 100 mL 1g = 1000mg c) convert g to mg!

Problem 12-5: Redox Titration - IV a) Mol Ca 2+ per 100 mL Blood b) mass (g) of Ca 2+ c) mass (mg) of Ca 2+

Answers to Problems in Lecture #12 1.(b) The ox. no. of oxygen is –2; the ox. no. of N is +4 (c) The ox. no. of H is +1, the ox. no. of each O is –2, the sulfur atom is +6 2.(b) S 8 is the reducing agent and O 2 is the oxidizing agent (c) CO is the reducing agent and NiO is the oxidizing agent 3.NH 3 + CH 4 -> HCN + 3H 2 4.3CuS + 8NO H 3 O+ -> 3Cu (SO 4 ) 2- +8NO + 12H 2 O 5.(a) 2.50 x mol Ca 2+ in H 2 SO 4 soln; 2.50 x mol Ca 2+ in blood; (b) 10.0 mg Ca 2+ /100 mL blood