Single Replacement Reactions

Transcription

1 Single Replacement Reactions CW single replacement reactions content a.doc As you have found out from your overview of chemical reactions, a single replacement reaction is one where an element reacts with a compound (often in aqueous solution) and an element and a compound are produced. A single replacement reaction is the chemical equivalent of "cutting in" at the chemical prom. Continuing our theme of studying the behavior of ions in solution, we will concentrate on single replacement reactions that occur in aqueous solution. If a reaction occurs and the element is a metal, then it will replace the metal in the compound. If a reaction occurs, and the element is a halogen, then it will replace the halogen in the compound. Zn(s) + Cu(NO 3 ) 2 (aq) Zn(NO 3 ) 2 (aq) + Cu(s) Cl 2 (aq) + 2NaI(aq) 2NaCl(aq) + I 2 (aq) Goals for your study of single replacement reactions: 1. Recognize a reaction as being single replacement. 2. Be able to predict if a reaction will occur or not (using the activity series). 3. Determine the formulas of the products (paying attention to oxidation numbers). 4. Determine if the compound that is formed is insoluble in water (solubility rules). 5. Balance the molecular equation. 6. Write a balanced net ionic equation. And just like some couples refuse to let someone cut in, some single replacement reactions simply don't occur. For instance, zinc metal will react with copper ions in solution... Zn(s) + Cu(NO 3 ) 2 (aq) Zn(NO 3 ) 2 (aq) + Cu(s) which means that zinc can "cut in", but copper metal won't react with zinc ions at all. Cu(s) + Zn(NO 3 ) 2 (aq) no reaction Zinc nitrate doesn't allow copper to "cut in". So just why do some single replacement reactions occur and others do not?

2 Goal One -- Pick out the single replacement reactions: 1) F 2 + NaBr NaF + Br 2 2) Ca(OH) 2 + Al 2 (SO 4 ) 3 CaSO 4 + Al(OH) 3 3) Mg + Fe 2 O 3 Fe + MgO 4) C 2 H 4 + O 2 CO 2 + H 2 O 5) PbSO 4 PbSO 3 + O 2 6) NH 3 + I 2 N 2 I 6 + H 2 7) H 2 O + SO 3 H 2 SO 4 8) H 2 SO 4 + NH 4 OH H 2 O + (NH 4 ) 2 SO 4 9) Na + H 2 O NaOH + H 2 10) H 2 + I 2 HI Did you find four reactions which are single replacement? List the numbers on the line below: Balance all of the reactions for practice, and identify the type of reaction for the remaining reactions. Do that on notebook paper. Goal Two -- Predict if a single replacement reaction actually will occur. Just like it is important to be able to determine if a double replacement reaction actually occurs when two compounds are mixed together by looking at the solubility rules, we must determine if a reaction occurs in a single replacement reaction. Will the element by itself replace a second element in a compound? To make that determination we use the activity series. You can find the activity series in your NCDPI reference tables on page 7. It is reprinted here for your convenience. Earlier, we mentioned the following two reactions. If a piece of zinc is added to a solution of copper(ii) nitrate the zinc will dissolve to form zinc(ii) ions (Zn 2+ ). The copper(ii) ions (Cu 2+ ) will come out of solution as copper metal and deposit on the surface of the zinc. Zn(s) + Cu(NO 3 ) 2 (aq) Zn(NO 3 ) 2 (aq) + Cu(s) If, instead, a piece of copper is added to a solution of zinc ions in solution, nothing will happen. The copper metal is not active enough to replace the zinc ions in solution. Cu(s) + Zn(NO 3 ) 2 (aq) no reaction We can predict those results by looking at the position of zinc and copper on the activity series. Zinc is well above copper, which is close to the bottom of the table. An element will replace the ions below it in the activity series, but not above it.

3 Look at the activity series for metals. Let s break the elements down by what they react with. Starting from the top we see alkali metals (Li, Na, K and Rb) plus the more active alkaline earth metals (Ca, Sr and Ba). These metals react easily with water. So simply adding any one of these metals to water will initiate a single replacement reaction making a solution of the metal hydroxide and liberating hydrogen gas. The Group IA metals make for some pretty exciting reactions, while the alkaline earth metals are much more tame. Look at the video: In English universities they have a person who conducts the demonstrations for lecture classes. I was the demonstrator when I was in high school. My first exposure to the reactions of sodium left an indelible mark on the ceiling of my chemistry room at our sister high school. Even without being told, I instinctively knew not to throw water on a sodium fire. For a small fee I will regale you with stories of sodium and potassium. Below are the molecular, ionic and net ionic equations. Notice that the ionic and net ionic equations are one in the same. There are no spectator ions. 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) 2Na(s) + 2H 2 O(l) 2Na + + 2OH - + H 2 (g) When an alkali metal is added to water which contains an acid-base indicator such as phenolphthalein, the solution will turn pink due to the excess of hydroxide ions, indicating that the solution is basic. As you can see from the photo, a great deal of heat is liberated, igniting the hydrogen gas that is produced. A problem for some teachers who may not be thinking it through completely, is to give a problem which contains an alkali metal and a metal ion which is lower in the activity series. For instance, I ve see this reaction proposed, and the student is asked to predict the products. The teacher thinks that Zn and NaCl(aq) will be the products. Na(s) + ZnCl 2 (aq)??? Does anyone see a problem with this? Sure, zinc is lower than sodium in the activity series, and some zinc metal may be produced, but the major reaction is going to be between sodium and the water in the solution. So this is a bad idea. You would not use sodium metal to recover the zinc in a solution of zinc chloride. The sodium will simply react with the water that is also present. So how would you convert zinc ions in a solution of zinc chloride to metallic zinc? Choose an element above zinc, but one that won t react with the water in the solution. Something like magnesium. Mg(s) + ZnCl 2 (aq) MgCl 2 (aq) + Zn(s) In this case the magnesium, being above zinc in the activity series will go into solution as magnesium ions, Mg 2+, and the zinc ions will be converted to elemental zinc metal. This brings us to the second group of metals in the activity series. The group of metals which react with water only when it is superheated steam consist of Mg, Al, Mn, Zn, Cr, Fe and Cd. These metals will react with hot water vapor to produce the metal ion and hydrogen gas. Also keep in mind that the elements at the top of the list, Li, Na, K, Rb, Ca, Sr and Ba, will also react with hot water vapor to produce the metal ion and hydrogen gas.

4 Al Al Passivating layer of Al 2 O 3 There is one interesting exception to this. Aluminum metal. Aluminum is an interesting metal because it appears to be very nonreactive. Your aluminum lawn furniture sits around outside for years and doesn t corrode. That s because aluminum is already corroded, and covered in a very thin layer of aluminum oxide, Al 2 O 3. Even when scratched off, the metal corrodes very quickly, reforming a layer of Al 2 O 3. The layer of aluminum oxide, Al 2 O 3, forms an impenetrable barrier of on the surface of aluminum. This barrier, called a passivating layer protects the aluminum from any further reaction with oxygen. But in a concentrated solution of base, the aluminum oxide is soluble and dissolves, exposing the bare aluminum metal. Under these conditions aluminum reacts vigorously with water, much like an alkaline earth metal, producing aluminum hydroxide and hydrogen gas. In an excess of hydroxide ions, the aluminum oxide layer dissolves to form a complex ion of aluminum and hydroxide ions. Al 2 O 3 + 2OH - + 3H 2 O 2[Al(OH) 4 ] - Now the aluminum metal is exposed, and it can react with water to make solid aluminum hydroxide and hydrogen gas. 2Al(s) + 6H 2 O 2Al(OH) 3 (s) + 3H 2 (g) This is the basis for the chemistry of solid drain cleaners like Drano. Drano consists of solid flakes of sodium hydroxide along with small pieces of aluminum foil. When wet, the hydroxide ions allow for the dissolving the passivating layer of aluminum oxide and the exposed aluminum metal reacts with water to make bubbles of H 2 gas. These bubbles of hydrogen gas provide the agitation needed to help the sodium hydroxide solution dissolve the hair that makes up the clog. Hair is soluble in concentrated base solutions, and hair is what is usually clogging the bathroom sink. Finally we see that all metals located above hydrogen in the activity series will react with acids (which produce hydrogen ions) to make hydrogen gas. In fact, the reaction consists of the metal reacting with the hydrogen ions from the acid. The anion of the acid is frequently a spectator ion. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) None of the metals below hydrogen will react with hydrogen ions to make hydrogen gas. This is why copper metal is impervious to almost all acids. The one exception, is nitric acid. Nitric acid, HNO 3, is the exception to the rule and is the only acid that many metals below hydrogen will dissolve in. But the key to these reactions is the nitrate ion. In the case of the vigorous reaction between copper and nitric acid, the reaction is actually between copper and the nitrate ion, NO 3 -. And this brings us back to Ira Remsen and the observation that nitric acid acts upon copper. 3Cu(s) + 8HNO 3 (aq) 3Cu(NO 3 ) 2 (aq) + 2NO(g) + 4H 2 O(l)

5 The reaction of copper with nitric acid is deceptive, in that the reactants look like a single replacement reaction. There is an element reacting with a compound. But the products don t fit the pattern. Therefore, the reaction of copper with nitric acid is more complex than a simple single replacement. Therefore, let us look at some single replacement reactions that do fit the pattern and apply your six goals. Goals for your study of single replacement reactions: 1. Recognize a reaction as being single replacement. 2. Be able to predict if a reaction will occur or not (using the activity series). 3. Determine the formulas of the products (paying attention to oxidation numbers). 4. Determine if the compound that is formed is insoluble in water (solubility rules). 5. Balance the molecular equation. 6. Write a balanced net ionic equation. The following reactions come from the worksheet taken from the internet and found at 1. Zinc metal is added to sulfur acid. We certainly have an element reacting with a compound. The reactants fit the pattern for a single replacement reaction where a metal could replace a metal ion or hydrogen ion. Zn(s) + H 2 SO 4 will zinc replace hydrogen ions? Since Zn is above H in the activity series, zinc will replace hydrogen and a reaction will occur. Go on to complete the remaining goals. What are the products? ZnSO 4 and H 2. Is ZnSO 4 soluble in water? Yes. It will be in aqueous solution. The other product is hydrogen gas, and hydrogen is diatomic, so write the formula as H 2 (g). The balanced molecular equation is: Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) Since zinc sulfate is soluble in water it will exist as Zn 2+ ions and SO 4 2- ions. But remember that sulfuric acid ionizes as H + and HSO 4 - ions. Therefore, the net ionic equation and the net ion will be the same: Zn(s) + H + + HSO 4 - Zn 2+ + SO H 2 (g) 2. Aluminum metal is added to a solution of (barely soluble) calcium sulfide (a poor choice on someone s part). It fits the pattern of a single replacement reaction, but since aluminum is below calcium, there will be no reaction. We would write: Al(s) + CaS(aq) no reaction 3. Chlorine gas is bubbled into a solution of potassium bromide. In this case a halogen reacts with the halide ion (bromide ion). Will chlorine replace bromine? Yes. Look at the activity series for halogens. Chlorine is above bromine. The element will replace an ion below it. Therefore, our balanced molecular equation is: Cl 2 (g) + 2KBr(aq) 2KCl(aq) + Br 2 (l) Pay attention to the state symbols and the solubility rules. KCl is soluble in water. Bromine and chlorine are both diatomic in the elemental state, so don t forget the subscript 2 for both of them. We would write the ionic equation as: Cl 2 (g) + 2K + + 2Br - 2K + + 2Cl - + Br 2 (l) Notice that potassium is a spectator ion, and so we can eliminate it from the net ionic equation: Cl 2 (g) + 2Br - 2Cl - + Br 2 (l)

6 4. Magnesium metal is placed into a solution of acetic acid (ethanoic acid). Again we have a metal reacting with a compound, in this case an acid. Since magnesium is above hydrogen in the activity series, a reaction will occur. The products will be a solution of magnesium acetate, which is soluble in water, and hydrogen gas. Mg(s) + 2HC 2 H 3 O 2 (aq) Mg(C 2 H 3 O 2 ) 2 (aq) + H 2 (g) Acetate ion will be the spectator ion, since both the acid and magnesium sulfate are soluble in water. Mg(s) + 2H + Mg 2+ + H 2 (g) 5. Chlorine gas is bubbled into a solution of magnesium bromide. This reaction is basically, a repeat of the reaction in #3, where the halogen will replace the halide ion. Remember that metals replace metals and halogens replace halogens. Cl 2 (g) + 2MgBr 2 (aq) 2MgCl 2 (aq) + Br 2 (l) Since magnesium ions are spectator ions, the net ionic equation is the same as it was in #3. Cl 2 (g) + 2Br - 2Cl - + Br 2 (l) 6. Potassium metal is added to a solution of magnesium bromide. Here is an example of a teacher picking an inappropriate set of reactants. Never mind whatever else is in solution, the potassium metal will react violently with the water in the solution. Some magnesium metal may be formed, and so we will write the reaction for that. Notice that we must write the formulas correctly in light of the oxidation numbers of Mg and K. K(s) + MgBr 2 (aq) 2KBr(aq) + Mg(s) Bromide ion is the spectator ion, and so the net ionic equation is: K(s) + Mg 2+ 2K + + Mg(s) Exercises: Refer to the six goals as you write the balanced molecular and net ionic equations for the following on notebook paper. Be sure to pay attention to writing the correct formulas of the products based on their oxidation numbers. Use the solubility rules to determine the state symbols. 1. Fe(s) + CuSO 4 (aq)??? Hint: iron in the product will be in the +2 oxidation state. 2. Rb(s) + H 2 O(l)??? Hint: This will make a big bang! 3. F 2 (g) + FeCl 3 (aq)??? 4. Zn(s) + H 2 S(aq)??? 5. Au(s) + H 2 SO 4 (aq)??? 6. Mg(s) + CrCl 3 (aq)??? 7. Mn(s) + AgNO 3 (aq)??? Hint: manganese in the product will be in the +3 oxidation state. 8. I 2 (aq) + NaCl(aq)??? 9. Ca(s) + H 2 O(l)??? 10. Co(s) + Fe(NO 3 ) 2 (aq)???