Oxidation and Reduction (REDOX) reactions? A reaction in which electrons are transferred from one atom to another is called an oxidation and reduction reaction or REDOX reaction.

Rules for Assigning Oxidation Numbers The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Free elements (uncombined state) have an oxidation number of zero. All Diatomics have an oxidation state of zero. Na, Be, K, Pb, H2, O2, P4 = 0 In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 The oxidation number of oxygen is –2. Exception: In H2O2 it is –1. 2

The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- Oxidation numbers of all the atoms in HCO3- ? 3

How to write oxidation numbers using rules?:Biochemhelp

Find the oxidation state of Cr in K2Cr2O7 Element K Cr O Total Charge of the compound has to equal ZERO Subscript Oxidation State of 1 atom Sum of oxidation states

Oxidation-Reduction Reactions “REDOX” reactions Determine the oxidation number of the boldface underlined element in the following formulas: NaClO4 NH4+ HCO3- 6

Concept Check Give the oxidation number for the following atoms: Cr3+ Cr = _____ O2 = _____ Cl-1 Cl = _____ Mg = _____ AgNO3 N = _____ NH3 N = _____

Concept Check Give the oxidation numbers for the atoms in (NH4)2CrO4 NH4NO3

How would you classify most metals and non- metals? Redox Reactions OXIDATION — loss of electron(s) by a substance. Increase in oxidation number. (Mg  Mg2+ + 2e-) REDUCTION — gain of electron(s) by a substance. Decrease in oxidation number. (S+ 2e-  S2-) OXIDIZING AGENT — electron acceptor and the substance is reduced. (Non-metals, Cl atom gains 1 e-) REDUCING AGENT — electron donor and the substance is oxidized. (Metals, Na atom loses 1 e-) How would you classify most metals and non- metals? 9

You can’t have one… without the other! Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons. You can’t have 2 oxidations or 2 reductions in the same equation. Reduction has to occur when there is oxidation. LEO the lion says GER! ose lectrons xidation ain lectrons eduction GER!

Another way to remember OIL RIG Oxidation is loss (oxidation number increases) Reduction is gain (oxidation number decreases)

Zn + 2HCl  ZnCl2 + H2 LEO OXIDATION Loss of e- Increase in oxidation number Reducing Agent GER REDUCTION Gain of e- Decrease or reduction in oxidation number Oxidizing agent Reaction of zinc with hydrochloric acid - YouTube Zn + 2HCl  ZnCl2 + H2

Examining Redox Reactions After oxidation numbers have been assigned, it can be determined that the reaction may or may not have been redox. Is this reaction a redox reaction? Explain why or why not. NaCl + AgNO3  AgCl + NaNO3

Redox Reactions Oxidation-Reduction Basics - YouTube 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Assign oxidation numbers to all atoms. Identify what is oxidized and what is reduced. Identify the oxidizing and reducing agent.

Concept Check Cu + 2HNO3  Cu(NO3)2 + H2 Give the oxidation number and identify what is oxidized, reduced, and identify the oxidizing and reducing agents. Cu + 2HNO3  Cu(NO3)2 + H2 16

Oxidation and Reduction half-reactions Shows either the oxidation or reduction part of a redox reaction. Shows the electrons gained or lost. Follow the law of conservation of matter: a) Mass of atom(s) on reactant side = mass of atom(s) on the product side. b) Conservation of charge: net charge must be the same on both sides of the equation, but does not necessarily equal zero.

Oxidation half-reaction Reduction half-reaction Half Reactions Oxidation half-reaction Shows an atom or ion losing 1 or more electrons while its oxidation number increases. Fe(s)  Fe3+(aq) + 3e- For oxidation, the e- is on the product side. Reduction half-reaction Shows an atom or ion gaining 1 or more electrons while its oxidation number decreases. Fe3+(aq) + 3e-  Fe(s) For reduction, the e- is on the reactant side.

Write the Half-reaction from an equation Cu(s) + AgNO3(aq)  Cu(NO3)2(aq) + Ag(s) Step 1: Assign an oxidation number to each element. Step 2: Write a partial half-reaction to show the change in oxidation state. (no e- yet) Step 3: Now show the number of e- needed to explain how the oxidation number changed. Step 4: Keep conservation of mass and charge for ½ reactions. Step 5: There must be a balance between the number of electrons lost and gained. Balance by multiplying through by the lowest common denominator. Step 6: Once the e- lost and gained are equal, we can cancel the e- on both sides and add the 2 half reactions.

A half reaction does not occur by itself. At least two such reactions must occur so that the electron released by one reactant is accepted by another in order to complete the reaction. Thus, oxidation and reduction reactions must take place simultaneously in a system. 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-) 19.1

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Half reactions Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Zn(s)  Zn2+(aq) + 2e- OXIDATION ½ reaction Cu2+(aq) + 2e-  Cu REDUCTION ½ reaction

Concept Check Balance the following reactions using the half-reaction method. Li(s) + CuSO4(aq)  LiSO4(aq) + Cu(s) Ag(s) + S(s)  Ag2S(s) AgS animation

The Galvanic or Voltaic Cell Zn/Cu Cu/Ag

Cell Construction KCl in agar Provides conduction between half-cells Salt bridge – KCl in agar Provides conduction between half-cells Observe the electrodes to see what is occurring. Cu Zn 1.0 M CuSO4 1.0 M ZnSO4

- + What about half-cell reactions? What about the sign of the electrodes? - + Why? cathode half-cell Cu+2 + 2e-  Cu anode half-cell Zn  Zn+2 + 2e- Cu plates out or deposits on electrode Zn electrode erodes or dissolves What happened at each electrode? Cu Zn 1.0 M CuSO4 1.0 M ZnSO4

Galvanic cell cathode half-cell (+) REDUCTION Cu+2 + 2e-  Cu anode half-cell (-) OXIDATION Zn  Zn+2 + 2e- overall cell reaction Zn + Cu+2  Zn+2 + Cu Spontaneous reaction that produces electrical current!

Standard Conditions Temperature - 25oC All solutions – 1.00 M Now for a standard cell composed of Cu/Cu+2 and Zn/Zn+2, what is the voltage produced by the reaction at 25oC? Standard Conditions Temperature - 25oC All solutions – 1.00 M All gases – 1.00 atm

- + Now replace the light bulb with a volt meter. 1.1 volts cathode half-cell Cu+2 + 2e-  Cu anode half-cell Zn  Zn+2 + 2e- Cu Zn 1.0 M CuSO4 1.0 M ZnSO4

We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE) H2 input 1.00 atm 25oC 1.00 M H+ 1.00 atm H2 Pt Half-cell 2H+ + 2e-  H2 inert metal EoSHE = 0.0 volts 1.00 M H+

Electrode - an electrical conductor which carries charge to or from a liquid undergoing electrolysis. Electrolysis Electrolysis is the break-down of a substance by electricity Electrolyte - a molten or aqueous solution through which an electrical current can flow. Lead bromide

Electrolysis experiments Electrolysis only happens in: - molten ionic liquids or - aqueous solutions containing ions. There must be a complete circuit. A lamp or ammeter shows that electricity is flowing around the circuit. This slide shows the basic equipment needed for electrolysis. If you have an interactive white board or the image is projected onto a normal white board, a student could be selected to label the diagram. A heat arrow could be added to show a molten electrolysis (which reiterates that ionic compounds have high melting points). No heat arrow is needed for solution electrolysis (which reiterates that many ionic compounds are soluble in water). It is important to reinforce that a circuit must be complete. This could be extended by a teacher drawing a wire between the 2 electrodes, which would make a short circuit, and posing the question as to whether electrolysis would happen. It would not, as the electricity would flow through the wire rather than the solution as the wire would have less resistance. It is vital that the electrodes are below the liquid surface, or again the circuit would not be complete. Please note that the circuit symbol for a lamp may be different, see your specification. Question to support this slide: Question 8 on page 117.

Electrolysis of zinc chloride

At the electrodes Anode (+) Cathode (-) (positive electrode) Negative ions go here (anions). As non-metal ions are negative, they go to the anode. Ions lose electrons. They are oxidised and become neutral atoms (which react together to form molecules). Cathode (-) (negative electrode) Positive ions go here (cations). As metal ions are positive, they go to the cathode. Ions gain electrons. They are reduced and become neutral atoms. This is an animated slide: when you click the left-hand mouse button once, the cathode statement will appear; click the button again and the contrasting anode statement will appear. This animation can be removed to have all text displayed by going into Slide Show, Custom Animation, then by highlighting each piece of text in turn and clicking (left-hand mouse key) Remove. This slide is colour-coded: red for anything which is positive (anode and cation); blue for anything which is negative (cathode and anion); and green for anything neutral. Questions to support this slide: Question 5 on page 117. Question 6 on page 117. Question 16 on page 138. Question 18 on page 139. Question 19 on page 139. Question 20 on page 139.