18 2 Balancing Oxidation Reduction Equations Cas 2

18. 2 Balancing Oxidation. Reduction Equations Ca(s) + 2 H 2 O(l) -----> Ca(OH)2 (aq) + H 2(g) 0 +1 +1 -2 0 reduction oxidation +2 -2

Vocabulary Oxidation- the loss of electrons Reduction- the gain of electrons LEO the lion goes GERLEO- Loses Electrons Oxidation GER- Gains Electrons Reduction Redox Reactions- (or oxidation-reduction reactions) A reaction that involves a transfer of electrons between two species. Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.

Rules For Balancing Redox Reactions 1) Both the mass (number of each type of atom) and the charge must be balanced 2) Elements (existing on their own, ex. Ca) are assigned an oxidation number of zero 3) Compounds have an overall charge of zero, so oxidation numbers of atoms within a compound sum to zero 4) Monatomic ions get the oxidation number of its charge (ex. Ca+ gets an oxidation number of 1+ 5) Polyatomic ions get a charge derived from the sum of the oxidation numbers on each atom (ex. Mn. O 4 - …. overall charge of 1 -, O almost always gets 2 -, so Mn needs a sign of 7+ to equal -1…. 7+(-2 x 4)=-1) 6) Group 1 elements are assigned 1+ and group 2 elements get 2+ 7) Oxygen usually gets 2 - and halogens usually get 1 -

Practice Assigning Oxidation Numbers 1) H 2 CO 3 H: +1, O: -2, C: +4 1) N 2 N: 0 1) overall charge = 0 Zn(OH)42 Zn: 2+, H: +1, O: -2 1) overall charge = 0 overall charge = -2 Li. H Li: +1, H: -1 overall charge = 0

Balancing Redox Reactions in Aqueous Solutions ● Use the half-reaction method of balancing ○ Break down the overall equation into two equations: one for oxidation and one for reduction ○ Balance each equation separately then add them together at the end ○ The steps for balancing reactions in an acidic solution differ from in a basic solution Ca(s) + 2 H 2 O(l) -----> Ca(OH)2 (aq) + H 2(g) 0 +1 +1 -2 0 Ca(s) -----> Ca(OH)2 (aq) oxidation 2 H 2 O(l) -----> H 2(g) reduction +2 -2

How To Balance Redox Reactions In Acidic Solutions Balance each half-reaction with respect to mass in the following order: 1. Balance all elements other than hydrogen and oxygen 2. Balance oxygen by adding water to the opposite side 3. Balance hydrogen by adding H+ to the opposite side

Practice Problem! Balance the redox reaction: Fe 2+(aq) + Mn. O 4 -(aq) ----> Fe 3+(aq) + Mn 2+(aq) Step 1: Assign oxidations states to all atoms and identify the substances being oxidized and reduced. Fe 2+(aq) + Mn. O 4 -(aq) ----> Fe 3+(aq) + Mn 2+(aq) +2 +7 -2 +3 Oxidized Reduced +2

Continued. . . Step 2: Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. Oxidations: Fe 2+(aq) ----> Fe 3+(aq) Reduction: Mn. O 4 -(aq) ----> Mn 2+(aq) Step 3: Balance each half-reaction with respect to mass (remember the order!!) Fe 2+(aq) ----> Fe 3+(aq) is already balanced with respect to mass 8 H+(aq) + Mn. O 4 -(aq) ----> Mn 2+(aq) + 4 H 2 O (l)

Continued………. . . Step 4: Balance each half-reaction with respect to charge by adding electrons. Fe 2+(aq) ----> Fe 3+ (aq) + 1 e 5 e- + 8 H+(aq) + Mn. O 4 -(aq) ----> Mn 2+(aq) + 4 H 2 O (l) Step 5: Make the number of electrons in both half-reactions equal by multiplying by a small whole number. 5[Fe 2+(aq) ----> Fe 3+ (aq) + 1 e- ] ===== 5 Fe 2+(aq) ----> 5 Fe 3+ (aq) + 5 e 5 e- + 8 H+(aq) + Mn. O 4 -(aq) ----> Mn 2+(aq) + 4 H 2 O (l)

Continued…………………. Step 6: Add the two half-reactions together and cancel the electrons. 5 Fe 2+(aq) ----> 5 Fe 3+ (aq) + 5 e 5 e- + 8 H+(aq) + Mn. O 4 -(aq) ----> Mn 2+(aq) + 4 H 2 O (l) 5 Fe 2+(aq) 8 H+(aq) + Mn. O 4 -(aq)----> 5 Fe 3+ (aq) + Mn 2+(aq) + 4 H 2 O (l)

Continued………………. . Step 7 (the final step!): Verify that the reaction is balanced both with respect to mass and with respect to charge. Reactants Products 5 Fe 2+(aq) 8 H+(aq) + Mn. O 4 -(aq) 5 Fe 3+ (aq) + Mn 2+(aq) + 4 H 2 O (l) 5 Fe 8 H 8 H 1 Mn +17 Charge

Practice Problem Balance the redox reaction in an acidic solution: Cu(s) + NO 3 -(aq) -----> Cu 2+(aq) + NO 2(g) Answer: Cu(g) + 4 H+(aq) + 2 NO 3 -(aq) -----> 2 NO 2(g) + 2 H 2 O(l) + Cu 2+(aq)

How to Balance Redox Reactions In Basic Solutions Balance each half-reaction with respect to mass in the following order: 1. 2. 3. 4. Balance all other elements other than hydrogen and oxygen Balance oxygen by adding water Balance hydrogen by adding H+ Neutralize H+ by adding enough OH- to neutralize each H+. Add the same number of OH- ions to each side of the equation.

Practice Problem! Balance the redox reaction: I-(aq) + Mn. O 4 -(aq) -----> I 2(aq) + Mn. O 2(s) Step 1: Assign oxidation states I-(aq) + Mn. O 4 -(aq)-----> I 2(aq) + Mn. O 2(s) -1 +7 -2 0 oxidized reduced +4 -2

Step 2: Separate into two half-reactions I-(aq) -----> I 2(aq) Mn. O 4 -(aq) -----> Mn. O 2(s) Step 3: Balance each half reaction 2 I-(aq) -----> I 2(aq) Mn. O 4 -(aq) -----> Mn. O 2(s) + 2 H 2 O 4 H+(aq) + Mn. O 4 -(aq) -----> Mn. O 2(s) + 2 H 2 O

4 H+(aq) + 4 OH- + Mn. O 4 -(aq) -----> Mn. O 2(s) + 2 H 2 O + 4 OH 4 H 2 O(l) + Mn. O 4 -(aq) -----> Mn. O 2(s) + 2 H 2 O + 4 OH- Step 4: Balance each half reaction with respect to charge 2 I-(aq) -----> I 2(aq) + 2 e 4 H 2 O(l) + Mn. O 4 -(aq) + 3 e- -----> Mn. O 2(s) + 2 H 2 O + 4 OH-

Step 5: Make the number of electrons equal 3[2 I-(aq) -----> I 2(aq) + 2 e-] 6 I-(aq) -----> 3 I 2(aq) + 6 e- 2[4 H 2 O(l) + Mn. O 4 -(aq) + 3 e- -----> Mn. O 2(s) + 2 H 2 O + 4 OH-] 8 H 2 O(l) + 2 Mn. O 4 -(aq) + 6 e- -----> 2 Mn. O 2(s) + 4 H 2 O + 8 OH-

Step 6: Add the half-reactions together 6 I-(aq) -----> 3 I 2(aq) + 6 e 4 8 H 2 O(l) + 2 Mn. O 4 -(aq) + 6 e- -----> 2 Mn. O 2(s) + 4 H 2 O + 8 OH- 6 I-(aq) + 4 H 2 O(l) + 2 Mn. O 4 -(aq) -----> 3 I 2(aq) + 2 Mn. O 2(s) + 8 OH-

Step 7: Verify! Reactants Products 6 I-(aq) + 4 H 2 O(l) + 2 Mn. O 4 -(aq) 3 I 2(aq) + 2 Mn. O 2(s) + 8 OH- 6 I 6 I 8 H 8 H 2 Mn 12 O -8 Charge

18. 3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions

Electrical Currents Electrical Current- a flow of electric charge Could be electrons going through a wire or ions going through a solution Redox reactions involve the transfer of electrons, so they have the potential to generate an electrical current! Let’s look at the following redox reaction to explain electrical currents: Zn(s) + Cu 2+(aq) -----> Zn 2+(aq) + Cu(s)

Zn(s) + Cu 2+(aq) -----> Zn 2+(aq) + Cu(s) Zn(s) Cu 2+(aq) Zn 2+(aq) Cu 2+(aq) Solid Zn loses electrons when placed in aqueous Cu 2+ (Zn is oxidized and Cu 2+ is reduced). On an atomic level, a zinc atom within zinc metal transfers two electrons to the copper ion solution. The zinc atom that transferred those two electrons then becomes a zinc ion dissolved in the solution. The copper ion that accepted the two electrons becomes solid copper on the zinc metal. What if we could separate the zinc atoms and the copper ions? If we connect solid zinc to the copper ions with a wire, the electrons would flow through the wire, creating an an electrical current. By doing so, we create a voltaic cell!

Electrochemical Cells Electrochemical Cell- A device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions. A voltaic cell (called Galvanic cell on the AP) is a type of electrochemical cell because it generates energy from a redox reaction. Another type of electrochemical cell is an electrolytic cell, which consumes electrical current to drive a nonspontaneous chemical reaction (this type of cell will be discussed later in the chapter).

Voltaic Cell

Voltaic Cells There are two half-cells, each with a metal strip placed into a solution. Connecting back to redox reactions, oxidation occurs in one cell and reduction occurs in the other. The metal strips in each half-cell (called electrodes) are connected with a wire. Electrons flow through the wire from the more negative electrode (called the anode, where oxidation occurs) to the positive electrode (called the cathode, where reduction occurs). This happens because the negative charge of the anode repels electrons. Remember: AN OX RED CAT Anode Oxidation Reduction Cathode Electrons always flow from the anode to the cathode

A solid strip of zinc is placed in Zn. SO 4 solution to form a half-cell. A solid strip of Cu is placed in Cu. SO 4 forms the second half-cell. The strips act as electrodes, or conductive surfaces through which electrons can enter or leave the half-cells. Each individual strip reaches equilibrium within its own solution. These half-reactions show this: Zn(s) <-----> Zn 2+(aq) + 2 e. Cu(s) <-----> Cu 2+(aq) + 2 e-

Zn(s) <-----> Zn 2+(aq) + 2 e. Cu(s) <-----> Cu 2+(aq) + 2 e. Zinc has a greater tendency to ionize, so its halfreaction lies further to the right. Because it lies to the right, the zinc electrode becomes negative as a result of the electrons from the forward reaction. The anode then repels electrons

As the electrons flow to the cathode, the Zn/Zn 2+ equilibrium shifts to the right again, leading to oxidation. The Cu/Cu 2+ equilibrium shifts to the left when it receives the electrons and reduction occurs. Copper ions then solidify on the metal strip. This is all according to Le Chatelier’s Principle!! Cu(s) <-----> Cu 2+(aq) + 2 e- More electrons = shift to the left

Salt Bridge Electrons go to the cathode, creating a buildup of negative charge. To even this charge out, the cell needs a pathway for counterions to flow. A salt bridge is the U-shaped bridge that connects the two half-cells. The bridge contains a strong electrolyte (Na. NO 3 in this example) and allows the flow of ions without allowing the solutions to mix. The positive ions in the bridge neutralize the negative ions in the cathodic solution that form from reduction. The negative ions in the bridge neutralize the positive ions in the anodic solution that form from oxidation.

Practice Problem! In a voltaic cell, electrons flow (a) From the more negatively charged electrode to the more positively charged electrode. (b) From the more positively charged electrode to the more negatively charged electrode. (c) From lower potential energy to higher potential energy. Answer: (a) From the more negatively charged electrode to the more positively charged electrode.

Units Electrical current is measured in amperes (A), also called amps. One amp represents the flow of one coulomb (a measure of electrical charge) per second 1 A = 1 C/s In addition, because an electron has a charge of 1. 602 X 10 -19 C: 1 A = 6. 242 X 1018 electrons per second

Units Electrical current is driven by a difference in potential energy (caused by an electric field resulting from the charge difference on the two electrodes) called potential difference, also known as electromotive force (emf) since it allows electrons to move. Potential difference is a measure of the difference in potential energy (in joules) per unit of charge (coulombs). The SI unit of potential difference is the volt (V), which is equal to one joule per coulomb. 1 V = 1 J/C ← in your reference table! Simple terms: a potential difference of 1 V means that a charge of 1 coulomb experiences an energy difference of 1 joule between two electrodes. Big potential difference = big difference in charge between two electrodes + strong tendency for electrons to flow

Units Potential difference between two electrodes in a voltaic cell is called the cell potential (Ecell) or cell emf. Cell potential depends on: 1) The relative tendencies of the reactants to undergo oxidation and reduction Strong tendencies result in a large potential difference 2) Concentrations of reactants and products and the temperature (we will assume 250 C unless otherwise stated) Under standard conditions (1 M concentration of reactants in solution and 1 atm pressure for gaseous products), the cell potential is called the standard cell potential (E 0 cell) or standard emf.

Units The cell potential is a measure of the tendency of a redox reaction to occur. The lower the cell potential, the lower the tendency to occur. A negative cell potential indicates that the forward reaction is not spontaneous.

Electrochemical Cell Notation Cell diagram (or line notation): A compact notation used to represent electrochemical cells and what is being oxidized and reduced Zn(s) | Zn 2+(aq) || Cu 2+(aq) | Cu(s) ● The oxidation half-reaction is on the left and the reduction is on the right ● The double vertical line represents the salt bridge separating the two reactions ● Substances in different phases are separated by a single line ● Some redox reactions have reactants and products in the same phase. In this case, we separate the reactants and products with a comma. ○ Ex. Fe(s) | Fe 2+(aq) || Mn. O 4 -(aq), H+, Mn 2+(aq) | Pt (s)