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CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART II Chapter Description 14 Chemical Equilibrium 15 Acids and Bases 16 Acid-Base Equilibrium 17 Solubility and Complex-Ion Equilibrium 19 Electrochemistry 20 Nuclear Chemistry

CH. 14 CHEMICAL EQUILIBRIUM DYNAMIC EQUILIBRIUM: WHEN FORWARD AND REVERSE REACTION RATES ARE THE SAME. *PURE SOLIDS AND LIQUIDS DO NOT APPEAR IN THE EQUILIBRIUM CONSTANT EXPRESSION (ANY K OR Q) THE EQUILIBRIUM CONSTANT K C : THE VALUE OBTAINED FOR THE EQUILIBRIUM-CONSTANT EXPRESSION WHEN EQUILIBRIUM CONCENTRATIONS ARE SUBSTITUTED. THE EQUILIBRIUM CONSTANT KP: THE EQUILIBRIUM CONSTANT EXPRESSION FOR A GASEOUS REACTION IN TERMS OF PARTIAL PRESSURES. FOR GASES: K P = K C (RT) ΔN, WHERE ΔN = DIFFERENCE OF COEFFICIENT. Ch. 14 Chemical Equilibrium

CALCULATING VALUES OF K IF 2 OR MORE REACTIONS ARE ADDED TO ACHIEVE A GIVEN REACTION, THE EQUILIBRIUM CONSTANT FOR THE GIVEN EQUATION EQUALS THE PRODUCT OF THE EQUILIBRIUM CONSTANTS (K) OF THE ADDED EQUATIONS. K 1 xk 2 IF REACTION IS REVERSED YOU TAKE THE INVERSE OF ORIGINAL K OF REACTION. 1/K IF MULTIPLIED OR DIVIDED, RAISE K TO THAT POWER. EX: MULTIPLY BY 2 EX: DIVIDE BY 3 K 2 K 1/3 Ch. 14 Chemical Equilibrium

USING Q C TO DETERMINE THE DIRECTION OF EQUILIBRIUM REACTION QUOTIENT (Q C ): THE INITIAL REACTION RATE OF A REACTION. ITS NOT A CONSTANT BUT DYNAMIC. THE RELATIONSHIP BETWEEN Q C AND K C GIVES THE DIRECTION OF THE EQUILIBRIUM. IF K C > Q C EQUILIBRIUM IS IN THE FORWARD DIRECTION. IF K C = Q C REACTION IS IN EQUILIBRIUM. IF K C < Q C EQUILIBRIUM IS IN THE REVERSE DIRECTION. IF Q C = 0 ONLY REACTANTS ARE PRESENT. IF Q C = ONLY PRODUCTS ARE PRESENT. Ch. 14 Chemical Equilibrium

1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide: N2O4(g) 2NO2(g) a) What is the value of Kc for this reaction? Kp = 0.660 at 319 K b) What is value of Kp for the reaction 2NO2 (g) N2O4(g) c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)? a) K p = K c (RT) Δn Kp K c = (RT) Δn 0.660 K c = (0.0821 x 319 ) 2-1 b) Reaction reversed then take the inverse of Kp 1/ K p = 1/ 0.660 c) (P NO2 ) 2 Kp = (P N 2O4 ) (0.223) 2 0.660 = (P N2O4 ) Ch. 14 Chemical Equilibrium

2. If a 2.50 L vessel at 1000 C containes 0.525 mol CO2, 1.25 mol CF4, and 0.75 mol COF2, in what direction will a net reaction occur to reach the equilibrium? CO2 (g) + CF4 (g) 2COF2 (g) Kc = 0.50 @ 1000 K Solution: How do you determine the direction of a reaction in order to reach equilibrium? [COF2] 2 Q c = [CO2] [CF4] [CO2] = 0.525 mol / 2.5 L = 0.21 M [CF4] = 1.25 mol / 2.5 L = 0.5 M [COF2] = 0.75 mol / 2.5 L = 0.3 M [0.3] 2 Q c = [0.21] [0.5] Kc = 0.50 Ch. 14 Chemical Equilibrium

3. Starting with 0.100 mol each of CO and H2O in a 5.00 L flask, equilibrium is established in the following reaction at 600K: CO(g) + H2O(g) CO2 (g) + H2 (g) Kc = 23.2 at 600K What is the concentration of hydrogen at equilibrium? Ch. 14 Chemical Equilibrium

Balanced Equation [CO] = 0.100 mol/ 5 L = 0.02 M [H2O] = 0.100 mol / 5 L = 0.02 M CO(g) + H2O(g) CO2 (g) H2 (g) Initial 0.02 0.02 0 0 Change - x -x +x +x Equilibrium 0.02-x 0.02-x x x [CO2] [H2] Kc = [CO] [H2O] [X] [X] 23.2 = [0.02-X] [0.02-X] Solution [X] 2 23.2 = [0.02-X] 2 *Take the square root of both sides: [X] 4.82 = [0.02-X] 0.096-4.82X = X 0.096 = X + 4.82X 5.82X = 0.096 X = 0.096/ 5.82 X = 0.0165 SO [H2]= X at equilibrium Ch. 14 Chemical Equilibrium

4. The K eq of a reaction is 4x10-7. At equilibrium. a) The products are favored b) The reactants are favored c) The reactants and products are present in equal amounts d) The rate of the forward and reverse reaction are the same Ch. 14 Chemical Equilibrium

4. The K eq of a reaction is 4x10-7. At equilibrium. a) The products are favored c) The reactants and products are present in equal amounts d) The rate of the forward and reverse reaction are the same Explanation: Here K eq = 0.0000004 K eq << 1 Therefore Reactants are favored!! What if K eq >> 1????? Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Shifts RX'N which way? Add PCl 5 Remove Cl 2 Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Add PCl 5 Shifts RX'N which way? Right Remove Cl 2 Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Add PCl 5 Remove Cl 2 Shifts RX'N which way? Right Right Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Add PCl 5 Remove Cl 2 Add Ar Shifts RX'N which way? Right Right No effect Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Add PCl 5 Remove Cl 2 Add Ar Decrease V (or increase P) Increase T Add Catalyst Shifts RX'N which way? Right Right No effect Left High to Low #of moles of Gases 1 mol 2 mol Right No effect on equilibrium *Catalysts Lowers Activation Energy only Ch. 14 Chemical Equilibrium

5. For the reaction, PCl 5 (g) PCl 3 (g) + Cl 2 (g) H rxn = +111 kj. Fill in the following table: Changes Add PCl 5 Remove Cl 2 Add Ar Decrease V (or increase P) Increase T Shifts RX'N which way? Right Right No effect Left High to Low #of moles of Gases 1 mol 2 mol Right Add Catalyst No effect on equilibrium *Catalysts Lowers Activation Energy only Which of the top changes would cause a change in K? ONLY a change in Temperature can cause a change in K Ch. 14 Chemical Equilibrium

Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH - ] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: Identify following as Bronsted-Lowry Acid/Base :NH 3 HCl

Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH - ] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: HCl (Acid) :NH 3 (Base) Lewis Definitions: 1) Acid: is an electron pair acceptor. 2) Base: is an electron pair donor. Example: Identify following as Lewis Acid/Base Ag + :NH 3

Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH - ] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: HCl (Acid) :NH 3 (Base) Lewis Definitions: 1) Acid: is an electron pair acceptor. 2) Base: is an electron pair donor. Example: Ag + (Acid) :NH 3 (Base)

STRONG ACIDS AND BASES HNO 3 H 2 SO 4 HClO4 GROUP 1A & 2A HYDROXIDE EX: NaOH, Ca(OH) 2 EXCEPT Be(OH) 2 HCl HBr HI H 3 O + or H + Ch. 15 Acids and Bases

COMPARING THE STRENGTHS OF GIVEN ACIDS Binary Acids (H-NONMETAL): A. Electronegativity if same period ( EN = strength) EN increases Left to Right B. Atomic size if same group ( atomic size = strength) Atomic size increases Top to Bottom Oxacids (H-O-nonmetal): Electronegativity ( EN = Strength) EN increases Left to Right and Bottom to Top EX: EX: HOCl vs HOBr EX: HI H 2 Te vs HI vs HF Other Groups (H-O-NM-O n ): More Oxygen = Stronger Acid EX: H 3 AsO 4 vs H 3 AsO 3 Carboxylic Acids (RCOOH): A. Resonance (EN) B. Inductive Effect: distance of EN element from COOH group. (Closer = More Acidic) EX: FCH 3 COOH vs FCH 3 CH 2 CH 2 COOH Ch. 15 Acids and Bases

Ch. 15 Acids and Bases Strength of Conjugate Acid and Conjugate Bases Strong base gives weak CA and weak base gives strong CA Strong acid gives weak CB and weak acid gives strong CB Reaction is favored in the direction which leads to the formation of weaker acid or base (or CB and CA). Ex: Which side of the following reaction is favored? CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

Strength of Conjugate Acid and Conjugate Bases Strong base gives weak CA and weak base gives strong CA Strong acid gives weak CB and weak acid gives strong CB Reaction is favored in the direction which leads to the formation of weaker acid or base (or CB and CA). Ex: Which side of the following reaction is favored? CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Answer: Weak Acid Strong C.B. Weak Base Strong C.A. Therefore the Reactant side (left) is Favored Ch. 15 Acids and Bases

1. Identify the acid and base on each side of the following equation: H 2 S + NH 3 NH 4 + + HS - Ch. 15 Acids and Bases

Ch. 15 Acids and Bases 1. Identify the acid and base on each side of the following equation: H 2 S + NH 3 NH 4 + + HS - ANSWER: Acid + Base C.A. + C.B.

2. Which of the following is the strongest Acid? A) HOI B) HOCl C) HOBr Ch. 15 Acids and Bases

Ch. 15 Acids and Bases 2. Which of the following is the strongest Acid? A) HOI C) HOBr Answer: HOCl With Oxacids (H-O-nonmetal) We use Electronegativity trend (increases bottom to top in a group) EN= Strength of Acid. EN

3. What is the ph of a solution prepared by dissolving 0.025 mol Ba(OH) 2 in water to give 455 ml of solution? Ch. 15 Acids and Bases

Ch. 15 Acids and Bases 3. What is the ph of a solution prepared by dissolving 0.025 mol Ba(OH) 2 in water to give 455 ml of solution? Solution: There is more than one way to solve this.. 0.025 mol x 2 = 0.05 mol OH - 0.05 mol/0.455 L = 0.1099 M OH - poh = -log [OH - ] poh = -log [0.1099] poh = 0.959 poh + ph = 14 ph = 14 - poh ph = 14 0.959

4. A sample of milk is found to have a ph of 6.25. What is the [OH - ] in this milk? Ch. 15 Acids and Bases

Ch. 15 Acids and Bases 4. A sample of milk is found to have a ph of 6.25. What is the [OH - ] in this milk? Solution: There is more than one way to solve this.. ph + poh = 14 poh = 14-6.25 poh = 7.75 [OH - ] = 10 -poh [OH - ] = 10-7.75 OR [H + ] = 10 -ph [H + ] = 10-6.25 [H + ] = 5.62x10-7 K w = [H + ] [OH - ] = 1x10-14 1x10-14 [OH - ] = = 5.62x10-7

CH. 16 ACID-BASE EQUILIBRIUM Acid ionization constant: *Lower value of pka = strong acid *Higher value of pka = weak acid Polyprotic acids: more than one ionizable proton e.g. H 2 SO 4 is diprotic and H 3 PO 4 is triprotic. In a solution there will be more than just the one conjugate acid/base pair. Base ionization constant: *Lower value of pk b = strong Base *Higher value of pk b = weak Base Ch. 16 Acid-Bases Equilibrium

Hydrolysis Strong Acid + Strong Base Neutral (ex. NaCl, KNO 3 ) Strong Acid + Weak Base Acidic Solution (ex. NH 4 Cl ) Weak Acid + Strong Base Basic Solution (ex. Na 2 CO 3 ) Weak Acid + Weak Base Depends on K and K b Ch. 16 Acid-Bases Equilibrium

Common Ion Effect The common ion in an equilibrium can shift the equilibrium to the opposite side. Ex: adding sodium acetate CH3COONa to the following reaction will increase the concentration of acetate ion thus moving the equilibrium to the left. Ch. 16 Acid-Bases Equilibrium

Buffers Buffers are solutions containing weak acid and its conjugate base OR weak base and its conjugate acid. The ph changes slightly with the addition of a little acid or base. Acid Base Indicators: chemicals that change color with ph. ph of buffers can be calculated using The traditional ICE method or by Henderson-Hasselbalch equation Ch. 16 Acid-Bases Equilibrium

Buffer Calculation: 1. What is the [H 3 O + ] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid K a = 5.80 x 10 7? Ch. 16 Acid-Bases Equilibrium

Buffer Calculation: 1. What is the [H 3 O + ] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid K a = 5.80 x 10 7? Solution: HA(aq) + H2O(l) H 3 O + (aq) + A (aq) pk a = log(5.80 x 10 7 ) = 6.237 [base] = 0.600 M and [acid] = 0.250 M Ch. 16 Acid-Bases Equilibrium

Ch. 17 Solubility Equilibrium CH. 17 Solubility and Complex-Ion Equilibrium The reactant in a solubility equilibrium is a slightly soluble salt and the equilibrium constant for the reaction is the Solubility Product Constant, K sp. Solubility problems are equilibrium problems. Ex: the solubility equilibrium and K sp for the salt SrF 2 (s) are SrF 2 (s) Sr 2+ (aq) + 2 F (aq) K sp = [Sr 2+ ][F ] 2 = 2.0 x 10-10 *Note that since the reactant is a Solid, its concentration Does Not appear in the K sp expression. The Molar Solubility of a salt in water can be formed by setting up an equilibrium table and solving for x. The Solubility is the same quantity expressed in g/l (rather than M = mol/l).

Ch. 17 Solubility Equilibrium Solubility Quotient (Q ip ) Solubility Quotient (Q ip ): determines if precipitation will occur Q ip > K sp precipitation occurs (Supersaturated) Q ip = K sp saturation of solution (Saturated) Q ip < K sp precipitation will not occur (Unsaturated) *In general very small value of K sp indicates complete precipitation)

Ch. 17 Solubility Equilibrium Factors Affecting Solubility Common ion effect: Addition of an ion common to a solubility equilibrium will reduce solubility. *Le Châtelier s Principle. Ex: SrF 2 (s) Sr 2+ (aq) + 2 F (aq) Adding fluoride ion, F (aq), to the SrF 2 (s) equilibrium above will shift it left. The shift will increase the amount of SrF 2 (s) in solid form, and thus decrease solubility. Effect of ph on solubility: ph of a solution will affect solubility if the conjugate ion (acid or base) is acidic or basic. Ex: Cl - is very weak conjugate base (CB) and is not considered basic, whereas HCO3 - is a weak CB whose solubility is affected by the ph. *An Anion (CB) of a weak Acid is more soluble in Acidic solutions.

Ch. 17 Solubility Equilibrium Comparing Precipitation Of Given Salts If two precipitates are possible, the least soluble salt will precipitate first (the one with the smaller K sp ). Ex: If you add NaOH(aq) to a solution containing equal amounts of Ca 2+ and Mg 2+ *Mg(OH) 2 (s) will precipitate before Ca(OH) 2 (s) since K sp (Mg(OH) 2 ) = 1.2 x 10-11 < K sp (Ca(OH) 2 ) = 8.0 x 10-6

Common Ion Effect: 1. What would be the molar solubility of SrF 2 (s) in a 0.10 M NaF(aq) solution? K sp = 2.0 x 10-10 Ch. 17 Solubility Equilibrium

Ch. 17 Solubility Equilibrium Common Ion Effect: 1. What would be the molar solubility of SrF 2 (s) in a 0.10 M NaF(aq) solution? K sp = 2.0 x 10-10 We set up an equilibrium (ICE) table, but now we have an initial concentration of fluoride ion:

Ch. 17 Solubility Equilibrium Common Ion Effect: 1. What would be the molar solubility of SrF 2 (s) in a 0.10 M NaF(aq) solution? K sp = 2.0 x 10-10 Balanced Equation SrF 2 (s) Sr 2+ (aq) + 2 F (aq) Initial (M) 0 0.10 Change (M) +x +2x Equilibrium (M) x (0.10 + 2x) We next use the equilibrium concentrations in the table and the K sp value given above, and solve for x. [0.10/ 2.0 x 10-10 ] = 500,000,000 >>>>1000 Note that since x is small, (0.10 + 2x) ~ 0.10

2. Which of the following salts is most soluble in pure water? A) HgS, K sp = 2.0 x 10-53 B) AgI, K sp = 8.5 x 10-17 C) PbI 2, K sp = 7.1 x 10-9 D) CuS, K sp = 8.7 x 10-36 E) ZnS, K sp =1.6 x 10 4 Ch. 17 Solubility Equilibrium

Ch. 17 Solubility Equilibrium 2. Which of the following salts is most soluble in pure water? A) HgS, K sp = 2.0 x 10-53 B) AgI, K sp = 8.5 x 10-17 C) PbI 2, K sp = 7.1 x 10-9 D) CuS, K sp = 8.7 x 10-36 E) ZnS, K sp =1.6 x 10 4 The higher the K sp, the more soluble it would be.

Ch. 17 Solubility Equilibrium 2. Which of the following salts is most soluble in pure water? A) HgS, K sp = 2.0 x 10-53 B) AgI, K sp = 8.5 x 10-17 D) CuS, K sp = 8.7 x 10-36 E) ZnS, K sp =1.6 x 10 4 The higher the K sp, the more soluble it would be.

CH. 19 Electrochemistry Galvanic cell is Spontaneous E cell > 0 G < 0 Ch. 19 Electrochemistry

Galvanic/Voltaic Cells Galvanic/Voltaic Cells: A cell that has a spontaneous redox reaction. The terminology of a cell are: Electrode : a metal piece at which the electrochemical reaction takes place. Anode: where oxidation (loss of e - ) occurs. (AN OX) Cathode: where reduction (gain of e - ) occurs. (RED CAT) Half-cell: the reduction or the oxidation part of the cell. Coulomb (C): the unit of electric charge. Volt (V): one joule per coulomb. Voltmeter: measures volts. Cell potential (E cell ): electro-potential difference that moves the electrons from the anode to cathode. *Electrons flow from Anode to Cathode (Mnemonic: A to C in alphabetical order) Galvanic cell is Spontaneous E cell > 0 G < 0 Ch. 19 Electrochemistry

Ch. 19 Electrochemistry Standard Electrode Potential: the tendency for reduction to occur at an electrode under the conditions of SHE. More More + - Reduction Potential = Oxidizing Agent = Reduction Reduction Potential = Reducing Agent = Oxidation *Reduction potential = potential of the reduction half reaction. *Oxidation potential = potential of the oxidation half reaction = reverse the sign of the reduction potential. Standard Cell Potential: E o cell = E o (cathode) E o (anode) E o cell = E o (right) E o (left)

Ch. 19 Electrochemistry Oxidation-Reduction Oxidation Losing Electrons Reducing Agent Increases In Charge Reduction Gaining Electrons Oxidizing Agent Decreases In Charge

1. Identify the following two reactions: A) 1 is oxidation, 2 is reduction B) 1 is reduction, 2 is oxidation C) Neither D) Both oxidation E) Both reduction 1) Cu(s) Cu 2+ (aq) + 2e - 2) 2 Ag + (aq) + 2e - 2 Ag(s) Ch. 19 Electrochemistry

1. Identify the following two reactions: 1) Cu(s) Cu 2+ (aq) + 2e - 2) 2 Ag + (aq) + 2e - 2 Ag(s) B) 1 is reduction, 2 is oxidation C) Neither D) Both oxidation E) Both reduction LeO GeR Ch. 19 Electrochemistry

2. The following questions are about this cell: Al Al +3 Pb +2 Pb a. Identify the anode and the cathode. b. Write the balanced overall reaction. c. What is the potential of this cell under standard conditions? Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation) Al loses e- and it is on the left, therefore it is the Anode Red Cat, GeR (Reduction Cathode, Gain e- Reduction) Pb gain e- and it is on the right, therefore it is the cathode

Ch. 19 Electrochemistry Oxidation- Reduction Reactions in Acidic/Basic Solutions Balance the following redox reactions (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. (b) give the balanced net reaction. (c) identify the oxidizing agent and the reducing agent. Cl 2 (g) + S 2 O 3 (aq) Cl - (aq) + SO4 (aq) in acid solution. 1. Assign Oxidation numbers and determine which is oxidized and which is reduced Cl2(g) + S 2 O 3 (aq) Cl (aq) + SO4(aq) Cl : 0-1 (reduced) S : +2 +6 (Oxidized) S 2 O 3 2S + 3O = 2S + 3() = 2S - 6 = 2S = +4 S = +2 SO 4 S + 4O = S + 4() = S - 8 = S = +6 S = +6 2. Split into two half reactions (Oxidation and Reduction) S 2 O 3 (aq) SO4 (aq) Cl 2 (g) Cl - (aq)

3. Balance both half Reactions 1) Balance all atoms except H and O S 2 O 3 (aq) 2 SO4 (aq) Cl 2 (g) 2 Cl - (aq) 2) Balance O atoms by adding H20 to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) 3) Balance H atoms by adding H + ions to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + 4) Balance electric charge by adding electrons (e-) to the more positive side S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- Cl 2 (g) + 2 e- 2 Cl - (aq) 4. Combine the two half reactions and obtain a final balanced equation a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- x4 (Cl 2 (g) + 2 e- 2 Cl - (aq)) 4 Cl 2 (g) + 8 e- 8 Cl - (aq) b. Simplify the balanced equation S 2 O 3 (aq) + 5 H20 + 4 Cl2 (g) 2 SO4 (aq) + 10 H + (aq) + 8 Cl - (aq) Ch. 19 Electrochemistry

3. Balance both half Reactions 1) Balance all atoms except H and O S 2 O 3 (aq) 2 SO4 (aq) Cl 2 (g) 2 Cl - (aq) 2) Balance O atoms by adding H20 to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) Oxidizing Agent? Reducing Agent? 3) Balance H atoms by adding H + ions to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + 4) Balance electric charge by adding electrons (e-) to the more positive side S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- Cl 2 (g) + 2 e- 2 Cl - (aq) 4. Combine the two half reactions and obtain a final balanced equation a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- x4 (Cl 2 (g) + 2 e- 2 Cl - (aq)) 4 Cl 2 (g) + 8 e- 8 Cl - (aq) b. Simplify the balanced equation S 2 O 3 (aq) + 5 H20 + 4 Cl2 (g) 2 SO4 (aq) + 10 H + (aq) + 8 Cl - (aq) Ch. 19 Electrochemistry

Ch. 19 Electrochemistry 3. Balance both half Reactions 1) Balance all atoms except H and O S 2 O 3 (aq) 2 SO4 (aq) Cl 2 (g) 2 Cl - (aq) 2) Balance O atoms by adding H20 to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) 3) Balance H atoms by adding H + ions to one side of the equation S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + 4) Balance electric charge by adding electrons (e-) to the more positive side S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- Oxidizing Agent? Reducing Agent? Cl 2 (g) S 2 O 3 (aq) Cl 2 (g) + 2 e- 2 Cl - (aq) 4. Combine the two half reactions and obtain a final balanced equation a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out S 2 O 3 (aq) + 5 H20 2 SO4 (aq) + 10 H + + 8 e- x4 (Cl 2 (g) + 2 e- 2 Cl - (aq)) 4 Cl 2 (g) + 8 e- 8 Cl - (aq) b. Simplify the balanced equation S 2 O 3 (aq) + 5 H20 + 4 Cl2 (g) 2 SO4 (aq) + 10 H + (aq) + 8 Cl - (aq)

If this was asking to balance in Basic solution, then there are two more steps. 5. Add as many OH - ions to both sides of the equation as there H + ions. S 2 O 3 (aq) + 5 H20 + 4 Cl2 (g) 2 SO4 (aq) + 10 H + (aq) + 8 Cl - (aq) + 10 OH - (aq) 6. each added OH - will react with H + to form H 2 O S 2 O 3 (aq) + 5 H2 0 + 4 Cl 2 (g) 2 SO4 (aq) + 10 H 2 0 + 8 Cl - (aq) *simplify the equation S 2 O 3 (aq) + 4 Cl2 (g) 2 SO4 (aq) + 5 H 2 0 + 8 Cl - (aq) Ch. 19 Electrochemistry

Bibliography Ebbing, D. D., & Gammon, S. D. (2013). General Chemistry. Cengage Learning. General Chemistry Course Information. (2014). Retrieved from Dr. Sapna Gupta: http://drsapnag.manusadventures.com/index.php/general-chemistry