CHEMISTRY CHEMI STRY CALCULATIONS CALCULATIONS LABORATORY
CHEM 113 3RD TERM AY 2015-2016 DEPT. OF CHEMICAL ENGINEERING UNIVERSITY OF SANTO TOMAS
REDUCTION-OXIDATION (REDOX) REACTIONS Redox reactions are chemical reaction involving transfer of electrons and change in oxidation numbers.
Oxidation number is a positive or negative number (or zero) assigned to an atom in a compound according to arbitrary rules that take into account bond polarity .
Oxidation number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element.
ARBITRARY RULES IN ASSIGNING OXIDATION NUMBERS 1. Any uncombined atom or any atom in a molecules of an element (free state) is assigned an oxidation number of zero. Ex. Fe, S, I2, Cl2, O2 2. The oxidation number of a monoatomic ion is the same as the charge of the ion. In their compounds, Group IA metals (Li, Na, K, Rb and Cs) always have oxidation numbers of +1 while Group IIA elements (Be, Mg, Ca, Sr and Ba) always have oxidation numbers of +2. 3. The sum of the oxidation numbers of the atoms in a compound is zero, since compounds are electrically neutral. Ex. HCl: 1(H) + 1(Cl) = 1(+1) + 1(-1) = 0 HClO3: 1(H) + 1(Cl) + 3(O) = 1(+1) + 1(+5) + 3(-2) = 0
ARBITRARY RULES IN ASSIGNING OXIDATION NUMBERS 4. The sum of the oxidation numbers of the atoms that constitute a polyatomic ion equals the charge on the ion . Ex. NH4+1: 1(N) + 4(H) = 1(-3) + 4(+1) = +1 5. The oxidation number of FLUORINE, the most electronegative element , is -1 in all fluorinecontaining compounds. 6. In most oxygen-containing compounds, the oxidation number of OXYGEN is -2. There are, however, a few exceptions:
a. In peroxides, the oxygen atom has an oxidation number of -1.
Ex. barium peroxide, BaO2
b. In the superoxide ion, O2-, each oxygen atom has an oxidation number of -1/2.
Ex. potassium superoxide, KO 2
c. In OF , oxygen has an oxidation number of +2
ARBITRARY RULES IN ASSIGNING OXIDATION NUMBERS 7. The oxidation number of hydrogen is +1 in all its compounds except for metallic hydrides (e.g. CaH2 and NaH) in which hydrogen has a -1 oxidation state. 8. In a combination of two non-metals (either a molecule or a polyatomic ion) the oxidation number of the more electronegative element is negative and equal to the charge on the common monoatomic ion of that element. Ex. In PCl3, the oxidation number of Cl is -1 and P is +3 In CS2, the oxidation number of S is -2 and C is +4
ARBITRARY RULES IN ASSIGNING OXIDATION NUMBERS Summary of Arbitrary Rules in Assigning Oxidation Numbers 1. Any free state atom = 0 2. Monoatomic ion: Group I A = +1, Group IIA = +2 … 3. Sum of oxidation numbers in a compound (electrically neutral) = 0 4. Sum of oxidation numbers in a polyatomic ion = charge on ion 5. F = -1 6. O = -2; except for peroxides (-1), superoxides (-1/2), OF2 (+2) 7. H = +1; except for metallic hydrides (-1) 8. For combination of two non-metals, oxidation no. of more electronegative element is
REDUCTION-OXIDATION REACTIONS Oxidation is any chemical change in which a substance loses electrons and thus, increases in oxidation state. Ex. Zn Zn2+ + 2 eReduction is a chemical change in which a substance gains electrons and thus, decreases in oxidation state. Ex. S + 2 e- S2The substance that gives up electrons is referred to as the reducing agent (RA) or reductant. The substance that takes up electrons is referred to as the oxidizing agent (OA) or oxidant. Ex. Zn + S ZnS Zn is the reducing agent (RA) and S is the oxidizing agent (OA)
ASSIGNING OXIDATION NUMBERS Exercise State the oxidation number of the underlined element. 1. U2Cl10
6. K2Cr 2O7
2. BiO+
7. MnO4-
3. Na6V10O28
8. NH2OH
4. K2SnO3
9. Mg3UO6
5. Ta6O18
10. CaH2
REDUCTION-OXIDATION REACTIONS For each of the following reactions, identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent 1. Zn + Cl2 ZnCl2 2. 2 ReCl5 + SbCl3 2 ReCl4 + SbCl5 3. Mg + CuCl2 MgCl2 + Cu 4. 2 NO + O2 2 NO2 5. WO3 + 3 H2 W + 3 H2O
BALANCING REDOX REACTIONS Oxidation- Number Method or Valence- Change Method 1. Assign oxidation numbers for each atom in the chemical reaction. 2. Identify the substance oxidized and the substance reduced. 3. Compute for the change in oxidation number for the substance oxidized and the substance reduced. 4. Multiply the changes in oxidation number (electrons gained and electrons lost) by certain factors to make the values equal. These factors multiplied to achieve equal changes in oxidation number will serve as the molar ratio of the substance reduced to the substance oxidized. 5. Apply the molar ratio of the substance reduced to the substance oxidized into the chemical equation and adjust accordingly until both sides of the equation are molecularly balanced.
OXIDATION-NUMBER METHOD K 2 C r 2 O 7 + H 2 O + S Ox. No.: +1 +6 -2
+1 -2
0
S O 2 + K O H + C r 2 O 3 +4 -2
+1-2+1
+3 -2
REDUCTION OXIDATION Change in Oxidation Number: Cr:
(+3) – (+6)
= |-3 x 4| = 12
S:
(+4) – (0)
= |+4 x 3| = 12
∴
Ratio of Cr red to Sox = 4:3
For every 4 atoms of Cr reduced, 3 atoms of S are oxidized.
OXIDATION-NUMBER METHOD ∴
For every 4 atoms of Cr reduced, 3 atoms of S are oxidized.
Since there are 2 atoms of Cr in K2Cr 2O7 and Cr 2O3, add a coefficient of 2 for both components then add a coefficient of 3 for S and SO2.
2 K 2 C r 2 O 7 + H 2 O + 3 S
3 S O 2 + K O H + 2 C r 2 O 3
Balance the chemical reaction accordingly by adding coefficients to the remaining components.
2 K 2 C r 2 O 7 + 2 H 2 O + 3 S
3 S O 2 + 4 K O H + 2 C r 2 O 3
OXIDATION-NUMBER METHOD 2 K 2 C r 2 O 7 + 2 H 2 O + 3 S
3 S O 2 + 4 K O H + 2 C r 2 O 3
Check if both sides of the equation is molecularly balanced. Reactant Side
Product Side
K
=2x2=4
K
=4x1=4
Cr
=2x2=4
Cr
=2x2=4
O
= 2(7) + 2(1) = 16
O
= 3(2) + 4(1) + 2(3) = 16
H
=2x2=4
H
=4x1=4
S
=3x1=3
S
=3x1=3
CLASSROOM ACTIVITY Balance the following molecular equations using the oxidationnumber method: 1. HNO3 + I2 NO2 + HIO3 + H2O 2. Bi(OH)3 + K2SnO2 Bi + K2SnO3 + H2O 3. HNO3 + HI NO + I2 + H2O 4. Na2Cr 2O7 + FeCl2 + HCl CrCl3 + FeCl3 + NaCl + H2O
BALANCING IONIC EQUATIONS BY ION-ELECTRON METHOD OR METHOD OF HALF-REACTIONS 1. Assign oxidation numbers for each atom in the chemical reaction. 2. Identify the substance oxidized and the substance reduced. 3. Establish a half -reaction for oxidation and another for reduction. 4. Balance each half -reaction molecularly by adding coefficients (if needed). To balance O, add H2O. To balance H, add H+. 5. Balance the charges of each half -reaction by adding electrons, e-. 6. Combine the two half -reactions such that the electrons cancel out.
BALANCING IONIC EQUATIONS BY ION-ELECTRON METHOD OR METHOD OF HALF-REACTIONS 7. Perform cancellation on any components that exists on both sides of the chemical reaction. (balancing stops here for acidic solutions) For reactions on basic or alkaline solutions: 8. Perform neutralization by adding OH- on both sides equal to the existing number of H + ions. 9. Combine the H+ and OH- ions on one side of the equation to form water, H2O (e.g. 2 H+ + 2 OH- = 2 H2O)
BALANCING IONIC EQUATIONS BY ION-ELECTRON METHOD OR METHOD OF HALF-REACTIONS M n O 4- + C l O 2Ox. No.
+7 -2
+3 -2
M n O 2 + C l O 4+4 -2
REDUCTION Half -reactions: Oxidation: ClO2- ClO4Reduction: MnO4- MnO2
OXIDATION
+7 -2
(base)
BALANCING HALF-REACTIONS
Balance each half -reaction molecularly by adding coefficients (if needed). To balance O, add H2O. To balance H, add H+.
Half -reactions: Oxidation: ClO2- ClO42 H2O + ClO2- ClO4- + 4 H+ Reduction: MnO4- MnO2 4
H+
+
MnO4-
MnO2 + 2 H2O
Add 2 molecules of H2O on the left side of the equation to balance O. There will be an excess of 4 atoms of H on the reactant side. To balance, add 4 H+ ions on the right side. Add 2 molecules of H2O on the right side of the equation to balance O. There will an excess of 4 atoms of H on the product side. To balance, add 4 H+ ions on the left side.
BALANCING HALF-REACTIONS
Balance the charges of each half reaction by adding electrons, e-.
Half -reactions: Oxidation: charge:
2 H2O + ClO2- ClO4- + 4 H+ 0
-1
0 + (-1) = -1
-1
-1 + 4 = +3
2 H2O + ClO2- ClO4- + 4 H+ + 4 e0
-1
0 + (-1) = -1
-1
4 (+1)
4(+1) 4(-1)
-1 + 4 + (-4) = -1
Determine the total charge of each side of the equation.
To balance the charges on both sides of the equation, add sufficient electrons to the more positive side.
BALANCING HALF-REACTIONS
Balance the charges of each half reaction by adding electrons, e-.
Half -reactions: Reduction:
4 H+ + MnO4- MnO2 + 2 H2O
Charge:
4(+1)
-1
4 + (-1) = +3
0
0
0+0=0
Determine the total charge of each side of the equation.
3 e- + 4 H+ + MnO4- MnO2 + 2 H2O To balance the charges on both sides of the 3(-1) 4(+1) -1 0 0 equation, add sufficient electrons to the more positive side. -3 + 4 + (-1) = 0 0+0=1
COMBINING BALANCED HALF-REACTIONS
Combine the two half -reactions such that the electrons cancel out.
Oxidation:
2 H2O + ClO2- ClO4- + 4 H+ + 4 e-
Reduction:
3 e- + 4 H+ + MnO4- MnO2 + 2 H2O
Multiply the half -reactions to factors that will make the number of respective electrons equal.
Oxidation:
( 2 H2O + ClO2- ClO4- + 4 H+ + 4 e- ) x 3
Reduction:
( 3 e- + 4 H+ + MnO4- MnO2 + 2 H2O ) x 4
COMBINING BALANCED HALF-REACTIONS
Combine the two half -reactions such that the electrons cancel out.
Oxidation:
6 H2O + 3 ClO2- 3 ClO4- + 12 H+ + 12 e-
Reduction:
12 e- + 16 H+ + 4 MnO4- 4 MnO2 + 8 H2O
6 H2O + 16 H+ + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 12 H+ + 8 H2O Perform cancellation
chemical reaction.
on any components that exists on both sides of the 8 – 6 H2O = 2 H2O (product side) 16 – 12 H+ = 4 H+ (reactant side)
4 H+ + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 2 H2O Note that balancing will end here for acidic solutions
NEUTRALIZATION (FOR BASIC/ALKALINE SOLUTIONS)
Perform neutralization by adding OH- on both sides equal to the existing number of H+ ions.
4 H+ + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 2 H2O No. of H+ ions = 4. Hence, 4 OH- ions should be added to both sides. 4 OH- + 4 H+ + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 2 H2O + 4 OH
Combine the H+ and OH- ions on one side of the equation to form water, H2O
4 OH- + 4 H+ = 4 H2O
4 – 2 H2O = 2 H2O (reactant side)
4 H2O + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 2 H2O + 4 OH2 H2O + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 4 OH-
BALANCING IONIC EQUATIONS 2 H2O + 3 ClO2- + 4 MnO4- 3 ClO4- + 4 MnO2 + 4 OHcharge: 2(0) 3(-1) 4(-1) 0 + (-3) + (-4) = -7 no. of atoms: reactant side H =2x2=4 O = 2(1) + 3(2) + 4(4) = 24 Cl = 3 x 1 = 3 Mn = 4 x 1 = 4
3(-1) 4(0) 4(-1) (-3) + 0 + (-4) = -7 product side H =4x1=4 O = 3(4) + 4(2) + 4(1) = 24 Cl =3x1=3 Mn = 4 x 1 = 4
CLASSROOM ACTIVITY Balance the following ionic equations using the ion -electron method 1. AsO2- + MnO4- AsO3- + Mn2+
(acid)
2. Cl2 ClO3- + Cl-
(basic)
CLASSROOM ACTIVITY Balance the following redox reactions 1. Bi(OH)3 + Na2SnO2 Na2SnO3 + Bi + H2O (using oxidation-number method) 2. S2- + NO3- S + NO
(acid)
(using ion-electron method)
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