Electrochemistry

Electrochemistry Electrochemistry involves either… generating electricity by harnessing a spontaneous chemical reaction (one with K > 1) OR using electricity to force a chemical reaction to occur (one that is non-spontaneous, K<1) Understanding electrochemistry requires a basic understanding of the processes called “oxidation” and “reduction”.

Oxidation States Oxidation states are numbers assigned to atoms that reflect the net charge an atom would have if the electrons in the chemical bonds involving that atom were assigned to the more electronegative atoms. Oxidation states can be thought of as “imaginary” charges. They are assigned according to the following set of rules:

Assigning Oxidation States Each atom in a pure element has an oxidation state of “0”. e.g. F2 O3 K N2 Fe For mono-atomic ions, the oxidation state is equal to the charge on the ion. e.g. F- Al3+ Fe2+ O2- The oxidation state of oxygen in a compound (e.g. CH3OH) is almost always “-2”. The oxidation state of hydrogen in a compound (e.g. CH3OH) is almost always “+1”. The sum of the oxidation states in a neutral compound must equal “0”. The sum of the oxidation states in a complex ion must equal the charge on the complex ion.

Examples - assigning oxidation numbers Assign oxidation states to all elements:

More Definitions Oxidizing Agent Reducing Agent the substance in a chemical reaction which causes another species to be oxidized. the oxidizing agent always gets reduced in the reaction. Reducing Agent the substance in a chemical reaction that causes another species to be reduced. the reducing agent always gets oxidized!

Examples - labeling redox reactions In each reaction, look for changes in oxidation state. If changes occur, identify the substance being reduced, and the substance being oxidized. Identify the oxidizing agent and the reducing agent. 5 Fe2+ + MnO4- + 8 H+ ® 5 Fe3+ + Mn2+ + 4 H2O H2 + CuO ® Cu + H2O Zn + 2 HCl ® ZnCl2 + H2

Example - Balancing with Half-Reactions Fe2+ + Cr2O72- ® Fe3+ + Cr3+ Fe2+ ® Fe3+ Fe2+ ® Fe3+ + e- 6 Fe2+ ® 6 Fe3+ + 6 e- Cr2O72- ® Cr3+ Cr2O72- ® 2 Cr3+ Cr2O72- ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ + 6 e- ® 2 Cr3+ + 7 H2O Cr2O72- + 6 Fe2+ + 14 H+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O

Example - Balancing with Half-Reactions Fe2+ + Cr2O72- ® Fe3+ + Cr3+ Fe2+ ® Fe3+ Fe2+ ® Fe3+ + e- 6 Fe2+ ® 6 Fe3+ + 6 e- Cr2O72- ® Cr3+ Cr2O72- ® 2 Cr3+ Cr2O72- ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ + 6 e- ® 2 Cr3+ + 7 H2O Cr2O72- + 6 Fe2+ + 14 H+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O

Example - Balancing with Half-Reactions Fe2+ + Cr2O72- ® Fe3+ + Cr3+ Fe2+ ® Fe3+ Fe2+ ® Fe3+ + e- 6 Fe2+ ® 6 Fe3+ + 6 e- Cr2O72- ® Cr3+ Cr2O72- ® 2 Cr3+ Cr2O72- ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ + 6 e- ® 2 Cr3+ + 7 H2O Cr2O72- + 6 Fe2+ + 14 H+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O

Example - Balancing with Half-Reactions Fe2+ + Cr2O72- ® Fe3+ + Cr3+ Fe2+ ® Fe3+ Fe2+ ® Fe3+ + e- 6 Fe2+ ® 6 Fe3+ + 6 e- Cr2O72- ® Cr3+ Cr2O72- ® 2 Cr3+ Cr2O72- ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ ® 2 Cr3+ + 7 H2O Cr2O72- + 14 H+ + 6 e- ® 2 Cr3+ + 7 H2O Cr2O72- + 6 Fe2+ + 14 H+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O

What if the solution was basic? Notice that the method has assumed the solution was acidic - we added H+ to balance the equation. The [H+] in a basic solution is very small. The [OH-] is much greater. For this reason, we will add enough OH- ions to both sides of the equation to neutralize the H+ in the overall reaction. The hydrogen and hydroxide ions will combine to make water, and you may have to do some canceling before you’re done. Cr2O72- + 6 Fe2+ + 7 H2O ® 2 Cr3+ + 6 Fe3+ + 14 OH- Why is this reaction unlikely in basic solution?

Balancing Redox Equations Practice Balance in acidic solution: H2C2O4 + MnO4- ® Mn2+ + CO2 5 H2C2O4 + 2 MnO4- +16 H+ ® 2 Mn2+ + 10 CO2 + 8 H2O Balance in basic solution: CN- + MnO4- ® CNO- + MnO2 3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

Balancing Redox Equations Practice Balance in acidic solution: H2C2O4 + MnO4- ® Mn2+ + CO2 5 H2C2O4 + 2 MnO4- +16 H+ ® 2 Mn2+ + 10 CO2 + 8 H2O Balance in basic solution: CN- + MnO4- ® CNO- + MnO2 3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

Balancing Redox Equations Practice Balance in acidic solution: H2C2O4 + MnO4- ® Mn2+ + CO2 5 H2C2O4 + 2 MnO4- +16 H+ ® 2 Mn2+ + 10 CO2 + 8 H2O Balance in basic solution: CN- + MnO4- ® CNO- + MnO2 3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

Balancing Redox Equations Practice Balance in acidic solution: H2C2O4 + MnO4- ® Mn2+ + CO2 5 H2C2O4 + 2 MnO4- +16 H+ ® 2 Mn2+ + 10 CO2 + 8 H2O Balance in basic solution: CN- + MnO4- ® CNO- + MnO2 3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

Balancing Redox Equations Practice Balance in acidic solution: H2C2O4 + MnO4- ® Mn2+ + CO2 5 H2C2O4 + 2 MnO4- +16 H+ ® 2 Mn2+ + 10 CO2 + 8 H2O Balance in basic solution: CN- + MnO4- ® CNO- + MnO2 3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

Redox Reactions - What’s Happening? Zinc is added to a blue solution of copper(II) sulfate The blue colour disappears…the zinc metal “dissolves”, and solid copper metal precipitates on the zinc strip The zinc is oxidized (loses electrons) The copper ions are reduced (gain electrons) Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s) Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

Copper ions (Cu2+) collide with the zinc metal surface A zinc atom (Zn) gives up two of its electrons to the copper ion The result is a neutral atom of Cu deposited on the zinc strip, and a Zn2+ ion released into the solution

Harnessing the Electricity During the spontaneous redox reaction, electrons flow from the zinc atoms to the copper ions. Zn ® Zn2+ + 2 e- (oxidation) Cu2+ + 2 e- ® Cu (reduction) Electricity can be thought of as the “flow of electric charge”. How can we make use of the flow of electrons from the zinc atoms to the copper ions?? Answer: SEPARATE the copper ions from the zinc atoms. This will force the electrons from the zinc atoms to travel through an external path to reach the copper ions.

Electrochemical Cells Definitions Voltaic/Galvanic Cell An electorchemical cell/ device that uses a spontaneous redox reaction to produce electricity Anode the electrode where oxidation occurs Cathode the electrode where reduction occurs Salt Bridge connects two “half-cells” to complete the electric circuit. for example, a U-tube filled with salt solution Electrochemical Cells Definitions

Electrochemical Cells

Cell Potential Water flows spontaneously over a waterfall because of a difference in potential energy between the top of the falls and the stream below. In a similar fashion, electrons flow from the anode of a voltaic cell to the cathode because of a difference in potential energy. The potential energy of electrons is higher in the anode than in the cathode, and they spontaneously flow through an external circuit from the anode to the cathode. The potential difference between the two electrodes of an electrochemical cell provides the driving force that pushes electrons through the external circuit. We call this potential difference, denoted Ecell, the cell potential. Because Ecell is measured in volts, we often refer to it as the cell voltage. For any cell reaction that proceeds spontaneously, such as that in a voltaic cell, the cell voltage will be positive.

Standard Hydrogen Electrode (SHE) It is impossible to measure the potential of a single electrode. Using a voltmeter, we can measure the difference in potential between two electrodes. Chemists arbitrarily assigned a potential of 0 V for the SHE. By measuring the difference in potential between the SHE and other electrodes, potentials can then be assigned to other electrodes. 2 H+ + 2 e- ® H2 E° = 0 V

When a zinc electrode is connected to the SHE as shown above, the voltmeter reads: E°cell = + 0.76 V The positive sign means electrons are flowing from zinc to the hydrogen cell. The zinc is being oxidized. Since E°SHE= 0 V, we can conclude that E°Zn = + 0.76 V . This is the “oxidation potential” for the reaction: Zn ® Zn2+ + 2 e-

Table of Standard Reduction Potentials

Calculating a Cell Potential Step 1: Which electrode is more likely to be reduced? This is the cathode. Step 2: Write half-reactions each half-cell - one reduction and one oxidation reaction. Include the values for E°red and E°ox from the table of standard reduction potentials. Remember: E°red = -E°ox Step 3: Calculate the cell potential: E°cell = E°red + E°ox

Step 1: Since Cu has a higher reduction potential than Zn, copper forms the cathode. We could also argue that Zn has a higher oxidation potential…and forms the anode. Step 2: Cathode: Cu2+ + 2e- ® Cu E°red = +0.34 V Anode: Zn ® Zn2+ + 2 e- E°ox = +0.76 V Step 3: The overall reaction is: Zn + Cu2+  Zn2+ + Cu E°cell = E°red + E°ox = 0.34 + 0.76 = 1.10 V Cu Zn

Zn | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu Line Notation An electrochemical cell needs to be described in a more convenient way than drawing a diagram! We use “line notation” to describe a cell. The zinc-copper standard cell is described: Zn | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu The ANODE is described before the CATHODE. Concentrations of ions are indicated in brackets. A vertical line represents a phase boundary. A double vertical line represents the salt bridge.

Line Notation - an example Calculate the cell potential for the cell described below: Mg | Mg2+ (1.0 M) || Ag+ (1.0 M) | Ag Anode: Mg ® Mg2+ + 2 e- E° = 2.37 V (oxidation) Cathode: Ag+ + e- ® Ag E° = 0.80 V (reduction) Overall: Mg + 2 Ag+  Mg2+ + 2 Ag E° = 3.17 V Review: What’s the oxidizing agent? the reducing agent? What’s E° for the reverse reaction?

Link to animation Click! And turn up the volume

The Nernst Equation So far, the cells we have considered have operated under standard conditions - 1.0 M solutions and 1.0 atm pressures at 25°C. Cell potential is dependent on concentration and on temperature, described by the Nernst Equation: where “E°“ is the standard cell potential “R” is the Ideal Gas Constant, 8.314 “T” is the temperature of the cell “n” is the number of moles of electrons transferred “F” is the Faraday, the charge on one mole of electrons, 96 500 C/mol “Q” is the reaction quotient (remember your equilibrium unit!)

Using the Nernst Equation Calculate Ecell for the cell described below at 25°C Mg | Mg2+ (0.0050 M) || Ag+ (2.0 M) | Ag According to the balanced equation, We have already seen that E°cell = 3.17 V According to our balanced equation, n = 2 mol e- Substituting into the Nernst Equation: Thus, Ecell = 3.26 V

Calculating Equilibrium Constants When an electrochemical cell operates, the concentrations of the ions change until Q = K . When the cell reaches equilibrium, Ecell = 0. Combining these facts with the Nernst equation gives: Rearranging, we can derive an equation to calculate the equilibrium constant for a redox reaction:

Calculating K for a Redox Reaction Calculate K for the reaction that occurs when Mg is added to a solution of AgNO3. The net reaction is: Mg + 2 Ag+  Mg2+ + 2 Ag

Calculating K for a Redox Reaction Calculate K for the reaction that occurs when Mg is added to a solution of AgNO3. The net reaction is: Mg + 2 Ag+  Mg2+ + 2 Ag 247 = ln (K) so K = e247 = HUGE

Corrosion Oxidation of metals with oxygen to form a metal oxide In some cases a thin layer of oxide coats the metal surface, preventing further oxidation (Al2O3) When iron corrodes, the process is called “rusting”. The iron oxide (Fe2O3) flakes off, exposing more metal to corrosion.

Galvanized Iron - Preventing Corrosion Iron can be protected from corrosion by coating it with zinc metal - a metal that will oxidize first, instead of the iron Note that E°oxidation of zinc is greater than that of iron.

Cathodic Protection - Preventing Corrosion To protect underground pipelines, a sacrificial anode is added. The water pipe is turned into the cathode and an active metal is used as the sacrificial anode. Magnesium is used as the sacrificial anode, since it is more easily oxidized than the iron water pipe.

Dry-Cell Batteries Zn + 2 MnO2 + 2 NH4+ ® Zn+2 + Mn2O3 + 2 NH3 + H2O

The Alkaline Battery Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s) (KOH is the electrolyte) Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)

Lead Storage Battery (Car Battery) Anode: Pb (s) + HSO4- (aq)  PbSO4 (s) + H+ (aq) + 2 e- Eo = 0.296 V Cathode: PbO2 (s) + 3 H+ (aq) + HSO4- (aq) + 2 e-  PbSO4(s) + 2 H2O () Eo = 1.628 V Overall: Pb (s) + PbO2 (s) + 3 H+ (aq) + 2HSO4  2 PbSO4(s) + 2 H2O () Eo = 1.924 V

A Picture of a Car Battery A 12-V battery has six cells connected in series. Each cell generates 2 V.

Fuel Cells Anode: 2 H2 (g) + 4 OH- (aq) 2 H2O () + 4 e- Cathode: O2 (g) + 2 H2O () + 4 e-  4 OH- (aq) Overall: 2 H2 (g) + O2 (g)  2 H2O ()

Electrolysis of Molten Sodium Chloride Anode: 2 Cl- ()  Cl2 (g) + 2 e- Cathode: 2 Na+ () + 2 e-  Na() Overall: 2 Cl- () 2 Na+ ()  Cl2 (g) + Na() Molten sodium metal is formed at the cathode. Chlorine gas is formed at the anode.

Electrolysis of Water Anode: 2 H2O  O2 + 4 H+ + 2 e- Cathode: 4 H2O + 4 e-  2 H2 + 4OH- Overall: 6 H2O  O2 + 2 H2 + 4 OH- + 4 H+ Since 4 OH- + 4 H+ ® 4 H2O Net: 2 H2O  O2 + 2 H2 H2 O2 H2O cathode anode Na2SO4

Electrolysis of Aqueous Sodium Bromide Two possible reactions at the cathode: 2 H2O + 2 e-  H2 +2OH- E° = -0.83 V Na+ + e- ® Na E° = -2.71 V Two possible reactions at the anode: 2 Br-  Br2 + 2 e- E° = -1.09 V 2 H2O ® O2 + 4 H+ + 4 e- E° = -1.23 V Overall reaction: 2 Br- + 2 H2O  Br2 + H2 + 2 OH- Eocell = Eored + Eoox = -0.83 + (-1.09) Eocell = -1.92 V

Quantitative Electrolysis Q = I t Charge = Current x time (Coulombs) (Amperes) (seconds) 1 F = 96 500 C/mol e-

What mass of aluminum can be produced in 8 What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3?

What mass of aluminum can be produced in 8 What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3? Al3+ + 3 e-  Al (s)

What mass of aluminum can be produced in 8 What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3? Al3+ + 3 e-  Al (s)

= 4.47 g of Aluminum will be produced What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3? Al3+ + 3 e-  Al (s) = 4.47 g of Aluminum will be produced