Electrochemistry Oxidation & Reduction Unit 16 Electrochemistry Oxidation & Reduction

Oxidation verses Reduction Gain oxygen atoms 2 Mg + O2  2 MgO Lose electrons (e-) Mg(s)  Mg+2 + 2 e- Lose hydrogen CH3OH  CH2O + H2 Lose oxygen atoms 2 KClO3  2 KCl + 3 O2 Gain electrons (e-) Cu+2 + 2 e-  Cu (s) Gain hydrogen CO + 2 H2  CH3OH

l l l l l l l l l -4 -3 -2 -1 0 1 2 3 4 reduction oxidation When oxidation occurs, there is also reduction. They go “hand in hand”. Not all reactions are oxidation reduction reactions.

Using Appendix 6 chart: To determine the strength as an oxidizing agent, look at the left side of appendix 6. The chemical that is higher on the list (has a greater magnitude + voltage) is the better oxidizing agent. Which is better? Br2 or Sn+2 To determine the strength as a reducing agent, look at the right side of appendix 6. The chemical that is lower on the list (has the greater magnitude – voltage) is the better reducing agent. Which is better? Pb(s) or H2O2

The chemical that is oxidized acts as the reducing agent The chemical that is oxidized acts as the reducing agent. The chemical that is reduced acts as the oxidizing agent. In the following, indicate which chemical is being oxidized, reduced, acts as the oxidizing agent, and acts as the reducing agent. Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)

Oxidation Reduction Lab Zn (s) Cu (s) Pb (s) Mg (s) Zn+2 Cu+2 Pb+2

Electricity is the movement of electrons to provide energy Electricity is the movement of electrons to provide energy. Two factors affect the amount of energy produced. 1. Voltage is the push behind the e-. Voltage is measured in volts 2. Current which is how many e- pass a point each second. Current is measured in Amperes. 1 Amp = 6.0 x 1018 e-/s An electrochemical Cell (Galvanic or voltaic cell) is a chemical device to make electricity from chemical reactions. It employs oxidation and reduction. It will have a positive (+) voltage while it works.

Cell Potential (Eocell) or Electromotive force (emf) is the pull or “driving force” on the e-. It is measured in volts (Eo ) Standard reduction potential is the potential for the half – reaction at 1M the standard being measured off 2 H+ + 2 e-  H2 , Pt electrode = 0.00 v Remember that these are listed as Reduction Potentials. When there is reduction, there is also oxidation. One of the two half – reactions will need to be reversed (so that it is written as an oxidation)

To calculate the voltage of an electrochemical cell: 1 To calculate the voltage of an electrochemical cell: 1. Obtain the two half – reactions from the reduction potential chart. 2. Determine which half – reaction is the oxidation, reverse the reaction. When the reaction is reversed, the sign on the voltage is changed. 3. Balance the e- in the two half – reactions by multiplying the coefficients by some whole number. Remember that voltage is a property of the kind of chemical, not the amount. Do not multiply the voltage. 4. Add the two half – reactions to determine the net voltage for the cell.

Oxidation Reduction Lab Zn (s) Cu (s) Pb (s) Mg (s) Zn+2 0.00 v No Yes Cu+2 Pb+2 0.00v

A salt bridge or porous disk connects the half cells and allows e- to flow, completing the circuit. We use a porous cup as the barrier between the 2 half cells

Calculate the voltage of a Galvanic cell made with Mg and Cu used to run the clock. Mg+2 + 2e-  Mg(s) volts Cu+2 + 2 e-  Cu(s) volts

Line Notation : A way to denote a galvanic cell Anode: the oxidation is written first Cathode: the reduction is written second One line separates the phases of a half cell Two lines separate the anode and cathode Mg l Mg+2 ll Cu+2 l Cu

Chemical Acid Base Neutral NaOH Na2Cr2O7 NaCl HCl H2O KOH Na2CO3 KNO3 HNO3 Na3PO4 Na2C2O4 NaSCN H2SO4 Corrosion of Iron Lab Check by each chemical if it is acid, base, or neutral

Corrosion is the process of returning metals to their natural state Corrosion is the process of returning metals to their natural state. The oxidation of metals. Example: 4 Fe(s) + 3 O2 (g)  2 Fe2O3 (s) All metals corrode easily, except Au, Ag, Cu Preventing corrosion: Coating with Cd or Sb Galvanizing (coating) with Zn Stainless steel is an alloy of Fe with Cd + Ni Sacrificial metals which corrode first are Mg or Zn

Corrosion Lab Fe (s)  Fe+2 + 2 e- Oxidation - Blue 2 e- + 2 H2O (L)  H2 (g) + 2 OH-1 Reduction - Pink Sacrificial metal Zn (s)  Zn+2 + 2 e- Oxidation - white

Corrosion of Iron Lab Check by each chemical if corrosion occurred Chemical Litmus Corrosion NaOH base Na2Cr2O7 base NaCl neutral HCl acid H2O neutral KOH base Na2CO3 base KNO3 neutral HNO3 acid Na3PO4 base Na2C2O4 base NaSCN base H2SO4 acid Corrosion of Iron Lab Check by each chemical if corrosion occurred

A battery is a group of galvanic cells connected in series A battery is a group of galvanic cells connected in series. (add potentials Eocell) Wet cell – contain fluid medium for e- flow Dry cell - contain a paste medium for e- flow Primary cell – one time use, not rechargeable Secondary cell - rechargeable

Dry Cell (Batteries) – do not contain a liquid electrolyte Acid version Anode reaction - oxidation Zn  Zn2+ + 2e +0.76 v Cathode reaction – reduction 2NH4+ + 2MnO2 + 2e  Mn2O3 + 2NH3 + 2H2O +0.74 v

2MnO2 + H2O + 2 e-  Mn2O3 + 2OH- cathode Other types Alkaline – Zn + 2OH-  ZnO + H2O + 2e- anode 2MnO2 + H2O + 2 e-  Mn2O3 + 2OH- cathode Silver cell – Zn anode, Ag2O cathode Mercury cell – Zn anode, HgO cathode Cadmium – anode, NiO2 – cathode (secondary – rechargable) Mercury cell

Pb + PbO2 + 2H2SO4  2PbSO4 + 2H2O +2.02 v Wet cell - Lead storage battery – rechargeable Anode Pb + HSO4-  PbSO4 + H+ + 2 e- +0.34 v Cathode PbO2 + HSO4- + 3 H+ + 2 e-  PbSO4 + 2 H2O +1.68 v Overall reaction Pb + PbO2 + 2H2SO4  2PbSO4 + 2H2O +2.02 v

An Electrolytic cell is a cell with a negative voltage, so there is no reaction. An outside source of energy is used as a way to drive the reaction. This type of cell is often used as a controlled way to deposit thin layers of metals, electroplating, on another metal.

Power source An electrolytic cell can be used to electroplate one metal on another metal. Pb Anode Cu Cathode Cr+3 + 3 e-  Cr Pb+2 + 2 e-  Pb Balance the reaction and determine the minimum voltage that needs to be supplied to drive the reaction.