Chapter 4 Reactions in Aqueous Solutions
Some typical kinds of chemical reactions: 1. Precipitation reactions: the formation of a salt of lower solubility causes the precipitation to occur. precipr 2. Acid Base reactions: the formation of water which is quite stable is a driving force for acid base chemistry. 3. Oxidation Reduction Reactions (reactions where electrons are gained and lost) the driving force for most reactions including oxidation reduction reactions is the drive to lower the potential energy of the system (that is to convert potential energy to kinetic energy usually in the form of heat) *cca1 glycerine; thermite Why do these reactions take place?
Electrolyte: the ability of a substance to form ions and conduct electricity Strong electrolytes: a substance when dissolved (usually in water) completely (or nearly so), dissociates into ions : MX +H 2 O = M + + X - +H 2 O ex. NaCl, HCl, H 2 SO 4 Weak electrolytes: a substance when dissolved (usually in water) partially dissociates into ions : MX +H 2 O = M + + X - + MX + H 2 O ex. acetic acid (vinegar, HC 2 H 3 O 2 ), HF Non-electrolytes: a substance when dissolved (usually in water) does not dissociate at all: MX +H 2 O = MX + H 2 O ex. sugar dvd MF:\Media_Assets\Chapter04\\StrongandWeakElectrolytes
Net Ionic Reactions Electrical neutrality requires the presence of both a cation (+) and anion to be present whenever we deal with any substance. However, some ions, either the cation (+) or anion (-) may merely be spectators in the chemical reaction that occurs but does not appear to play a role in the reaction. In writing net ionic reactions, these ions are removed from the equation if they do not undergo any significant change as a result of the reaction. Example: NaOH(aq) + HCl(aq) = NaCl (aq) + H 2 O Na + + OH - + H + + Cl - = Na + + Cl - + H 2 O nie (net ionic eq) OH - + H + = H 2 O
Example: Bi(OH) 3 + HCl(aq) = BiCl 3 + H 2 O Bi(OH) 3 + 3HCl(aq) = BiCl 3 + 3H 2 O Bi(OH) 3 (s) + 3H + Bi(OH) 3 (s) + 3H + + 3Cl - = Bi +3 + 3Cl - + 3H 2 O = Bi +3 + 3H 2 O What do you need to know? 1. Bi(OH) 3 is an insoluble precipitate 2. HCl and BiCl 3 are strong electrolytes 3. H 2 O is also mainly in the form of H 2 O, not H + + OH -
Solubility Rules: 1. Cations: a compound is probably soluble if it contains the following cations: alkaki metals ( Li +, Na +, K +, Rb + Cs+). 2. Anions: a compound is probably soluble if it contains the following anions: halogens (except for Ag +, Pb +2, and Hg 2 +2 ) nitrate (NO 3- ), perchlorate (ClO 4- ), acetate(c 2 H 3 O 2- ), sulfate (SO 4-2 ) (except Ba +2, Hg 2 +2 and Pb +2 sulfates. Most other cation-anion combinations form insoluble salts. Most soluble salts are strong electrolytes
Balance and write net ionic equations in water for each of the following: NiCl 2 + NaOH = Ni(OH) 2 + NaCl NiCl 2 + 2NaOH = Ni(OH) 2 + 2NaCl Ni +2 + 2 OH - = Ni(OH) 2 AlCl 3 + NaOH = NaAl(OH) 4 + NaCl AlCl 3 + 4NaOH = NaAl(OH) 4 + 3NaCl Al +3 + 4OH - = Al(OH) - 4 KOH + HC 2 H 3 O 2 (vinegar) = KC 2 H 3 O + 2 H 2 O OH - + HC 2 H 3 O 2 (vinegar) = C 2 H 3 O - 2 + H 2 O AgNO 3 + NaCl = AgCl + NaNO 3 Ag + + Cl - = AgCl
Solubility Rules: 1. Cations: a compound is probably soluble if it contains the following cation: alkaki metals ( Li +, Na +, K +, Rb + Cs + ). 2. Anions: a compound is probably soluble if it contains the following anions: halogens (except for Ag +, Pb +2, and Hg 2 +2 ) nitrate (NO 3- ), perchlorate (ClO 4- ), acetate(c 2 H 3 O 2- ), sulfate (SO 4-2 ) (except Ba +2, Hg 2 +2 and Pb +2 sulfates). Most other cation-anion combinations form insoluble salts. Are the following soluble? K 2 CrO 4 ZnCl 2 Pb(NO 3 ) 2 Ag 2 SO 4 Ca(NO 3 ) 2 BaS HgS Na 2 S
Relative reactivity of metals Na + H 2 O = NaOH + H 2 ; this reaction occurs with the elements in first two columns and with Al, Mn, Zn, Co, Ni, Sn. How can we determine which is most and which is least reactive?
Metals Highly Active Potassium, K Lithium, Li Barium, Ba Calcium, Ca Sodium, Na Magnesium, Mg Aluminum, Al Zinc, Zn Iron, Fe Nickel, Ni Tin, Sn Lead, Pb Hydrogen, H 2 Copper, Cu Mercury, Hg Silver, Ag M + H 2 O 2MOH +H 2 M + + H 2 M + 2 H+
2Al(s) + 3ZnCl 2 (l) = 2AlCl 3 (l) + 3Zn(s)? Metals Highly Active Potassium, K Lithium, Li Barium, Ba Calcium, Ca Sodium, Na Magnesium, Mg Aluminum, Al Zinc, Zn Iron, Fe Nickel, Ni Tin, Sn Lead, Pb Hydrogen, H 2 Copper, Cu Mercury, Hg Silver, Ag
Oxidation and Reduction Oxidation: the process by which an element or group of elements loose electrons Reduction: the process by which an element or group of elements gain electrons What is meant by the term agent? a facilitator In order to maintain electrical neutrality, for every electron lost by an element, there must be a gain of an electron by some other reactant. The oxidizing agent is the agent responsible for the loss of electrons. In the process the oxidizing agent get reduced The agent that looses electrons causes something else to gain electrons and therefore is the agent responsible for reduction Oxidizing agent is reduced Reducing agent is oxidized
How do I identify an oxidation reduction reaction? 1. Rules in Assigning oxidation states: All elements are in an oxidation state = 0 Metals usually get oxidized (become cations), non-metals usually get reduced (become anions) Typical oxidation states Alkali metals +1 Halogens -1 Alkaline earths +2 Group 6A -2 Group 3A +3 Group 5A -3 H can be 1 or +1
Some typical oxidation reduction reactions 1. Oxidation of paper : C 6 H 12 O 6 + 6O 2 = 6CO 2 + 6H 2 O 2. KMnO 4 + C 3 H 8 O 3 = CO 2 + Mn 2 O 3 + K 2 CO 3 3. 2Al + Fe 2 O 3 = Al 2 O 3 + 2 Fe How do we know that in these reactions, there have been loss and gain of electrons?
2Al + Fe 2 O 3 = Al 2 O 3 + 2 Fe Aluminum metal is neutral In Al 2 O 3, Al = +3 Iron metal is neutral In Fe 2 O 3, Fe = +3 Oxygen remains -2 on both sides Notice that this reaction could be balanced by mass balance alone.
What is the problem balancing oxidation-reduction reactions by mass balance only? Let balance this reaction only with regards to mass Cu + HNO 3 (aq) NO 2 (g) + H 2 O +Cu(NO 3 ) 2 (aq) Cu + 3HNO 3 = Cu(NO 3 ) 2 + NO 2 + H 2 O + H + A reaction that creates or destroys charge needs to be balanced by taking into account electron balance as well as mass balance. How do you know if mass balancing will not work? Charge will be created or destroyed by mass balance alone or you will need to form other products to balance the reaction
Balancing Oxidation and Reduction Reactions Two steps are involved in balancing oxidation-reduction reactions. Step 1: First, it is important to balance the loss and gain in electrons Step 2: Second, it is important to achieve mass balance How do I identify an oxidation reduction reaction that requires both charge and mass balance? If charge is created or destroyed when you mass balance an equation, or other products need to be produced, then you have an oxidation reduction equation that requires balancing both charge and mass
Balancing Oxidation and Reduction Reactions Two steps are involved in balancing oxidation-reduction reactions when the charge on either side of the equation is uneven. Step 1: First, it is important to balance the loss and gain in electrons Step 2: Second, it is important to achieve mass balance What do I do first? 1. Assign oxidation numbers
Assign oxidation states for each of the element in the following: H 2 SO 4 H = +1; O = -2; S = +6 Cl 2 0 H 3 PO 4 H = +1; O = -2; P = +5 HClO 4 H = +1; O = -2; Cl = +7 ZnS Zn = +2; S = -2 HNO 3 H = +1; O = -2; N = +5 Cr 2 O -2 7 Cr = +6; O =-2 MnO 4-1 Mn = +7; O = -2 MnO 2 Mn = +4; O = -2 C 6 H 12 O 6 C = 0; H = +1; O = -2 What is important in assigning oxidation states is not what oxidation state you give an element but how that state for the element compares on both side of the equation. An element with the same oxidation state on both sides of the equation is not involved in electron transfer and can be ignored H 2 O 2 H = +1; O = -1
Balancing Oxidation and Reduction Reactions Two steps are involved in balancing oxidation-reduction reactions. Step 1: First, it is important to balance the loss and gain in electrons Step 2: Second, it is important to achieve mass balance What do I do first? 1. Assign oxidation numbers 2. Determine what is oxidized and what is reduced
Which is the reducing agent (what is oxidized)? Cu + HNO 3 = Cu(NO 3 ) 2 + NO 2 Cu = 0; Cu +2 Cu is oxidized; Cu Cu +2 + 2e - (note that the reaction is overall neutral) Which is the oxidizing agent (gets reduced)? In HNO 3, N = +5 ; NO 2, N = +4 e - + HNO 3 NO + 2 OH - ( the reaction is overall neutral) N is reduced; note that the charge on oxygen is still -2, hydrogen is still +1
Balancing Oxidation and Reduction Reactions Two steps are involved in balancing oxidation-reduction reactions. Step 1: First, it is important to balance the loss and gain in electrons Step 2: Second, it is important to achieve mass balance What do I do first? 1. Assign oxidation numbers 2. Determine what is oxidized and what is reduced 3. Mass balance the oxidation half reaction; mass balance the reduction half reaction 4. Combine the two half reactions; for reactions taking place in H 2 O, it is permissible to break up water to form OH - and H + as necessary or to form water from OH - and H +.
Cu + HNO 3 = Cu(NO 3 ) 2 + NO 2 Cu Cu +2 + 2e - e - + HNO 3 NO 2 + OH - Cu Cu +2 + 2e - 2e - + 2HNO 3 2NO 2 + 2OH - Cu + 2HNO 3 2NO 2 + Cu(OH) 2 In acid medium: Cu(OH) 2 + 2 HNO 3 = Cu(NO 3 ) 2 + 2 H 2 O So: Cu +4 HNO 3 = Cu(NO 3 ) 2 + 2 H 2 O + 2 NO 2
Balanced equation Cu(s) + 4HNO 3 (aq) 2NO 2 (g) + 2H 2 O(l) +Cu(NO 3 ) 2 (aq) Net ionic equation Cu (s); NO 2 (gas) HNO 3 exists in H 2 O as H + + NO - 3 Cu(NO 3 ) 2 is a strong electrolyte and water soluble Cu(s) + 4H + (aq) + 2NO 3- (aq) 2NO 2 (g) + 2H 2 O(l) + Cu +2 (aq)
Balance the following equations: Fe(CN) -3 6 (aq) + N 2 H 4 (aq) = Fe(CN) -4 6 (aq) + N 2 (g) Oxidation H = +1; N in N 2 H 4 is -2; in N 2 : N = 0 Reduction Fe in Fe(CN) -3 6 = +3; Fe(CN) -4 6 = +2 N 2 H 4 (aq) = N 2 (g) + 4 H + + 4 e - 4e - + 4Fe(CN) 6-3 = 4Fe(CN) 6-4 N 2 H 4 (aq) + 4Fe(CN) 6-3 (aq) = 4Fe(CN) 6-4 (aq) + N 2 (g) + 4 H + (aq)
Let s look at an oxidation reduction reaction that can be balanced by mass balance CH 4 + 2O 2 = CO 2 + 2H 2 O Let s assign H as H +1 then C is C -4 both in CH 4 and in C +4 in CO 2 CH 4 = CO 2 + 8 e -1 + 4H + 2O 2 0 + 8 e -1 = 4O -2 CH 4 + 2O 2 = CO 2 + 4H + +2O -2 = CO 2 + 2 H 2 O If we assign H as H -1, then C is C +4 in both CH 4 and CO 2 C +4 H 4 = C +4 + 4H + +8 e -1 4H -1 goes to 4H +1 + 8 e -1 2O 2 + 8 e - = 2O -2 CH 4 + 2O 2 = CO 2 + 2H 2 O
PbO 2 (s) + Mn +2 (aq) = Pb +2 (aq) + MnO 4- (aq) in acid solution PbO 2, Pb = +4; Pb +2 reduction What s the half reaction? 2e - + PbO 2 (s) = Pb +2 (aq) +2O -2 Mn +2 ; MnO 4-, Mn = +7; oxidation What s the half reaction? 4O -2 + Mn +2 (aq) = MnO 4- + 5 e - 10e - + 5PbO 2 (s) = 5Pb +2 (aq) + 10 O -2 8O -2 + 2Mn +2 (aq) = 2MnO 4- + 10e - 5PbO 2 (s) + 2Mn +2 (aq) = 2MnO 4- + 5Pb +2 + 2O -2 4H + (aq) + 5PbO 2 (s) + 2Mn +2 (aq) = 2MnO 4- (aq) + 5Pb +2 (aq) + 2H 2 O
KMnO 4 + C 3 H 8 O 3 = CO 2 + Mn 2 O 3 + K 2 CO 3 + H 2 O MnO 4-, Mn = +7 Mn 2 O 3, Mn = +3 8e - + 2KMnO 4 Mn 2 O 3 + 5O-2 + 2K + C 3 H 8 O 3, C = -2/3; CO 2, C = +4 1C -2/3 = 1C +4 + 4.666 e - ; 3 C -2/3 = 3C +4 + 14e - 3O -2 + C 3 H 8 O 3 3CO 2 + 8H + + 14e - multiply 14*4 = 56; 7*8 = 56 56e - + 14KMnO 4 7Mn 2 O 3 + 35O -2 + 14K + 12O -2 + 4C 3 H 8 O 3 12CO 2 + 32H + + 56e - 14 KMnO 4 + 4C 3 H 8 O 3 12CO 2 +14K + +32H + +7Mn 2 O 3 + 23O -2 + 14KMnO 4 +4C 3 H 8 O 3 12CO 2 +7K 2 O +16H 2 O+7Mn 2 O 3 14KMnO 4 +4C 3 H 8 O 3 5CO 2 +7K 2 CO 3 +16H 2 O +7Mn 2 O 3