Essential Question: Why are chemical reactions with

Oxygen (REDOX) interdependent?

UNIT 10: OXIDATION/

REDUCTION

What does the chemical reaction between Magnesium & Oxygen look like? ¡ Mg = 2 e- & O = 6 e- ��

¡ Mg loses it’s TWO and O �gains them

¡ MgO is now formed

WHAT IS REDOX??

NOW YOU TRY: Magnesium and Chlorine? ¡ Mg = 2 e- & Cl = 7 e-

¡ Mg loses it’s TWO and EACH Cl gains ONE e- ¡ MgCl2 is now formed

WHAT IS REDOX??

OXIDATION Whenever an atom or ion

LOSES electrons

REDUCTION

These REDOX reactions NEVER take place alone…whenever one occurs, the

other MUST as well

Whenever an atom or ion

GAINS electrons

WHAT IS REDOX??

But how many electrons were exchanged? ¡ Assign Oxidation #s (states) à find on PT ¡ Oxidation = Loss of Electrons/Gain in

oxidation # ¡ Reduction = Gain of Electrons/Loss in

oxidation #

LEO says GER LEO = Loss of Electrons is Oxidation GER = Gain of Electrons is Reduction

OXIDATION #’S

RULES to assigning Oxidation #s: 1.  ALL “stand alone” elements (not in

compounds) have an oxidation # = ZERO ex) 2Na + Cl2 à 2NaCl àNa & Cl2 = oxidation #s of each = 0

2.  Monoatomic ions have oxidation # equal to the ionic charge

ex) NaCl à Na is +1 and Cl is -1 3.  Group 1 metals = �

Group 2 metals =

OXIDATION #’S

+1 Oxidation # +2 Oxidation #

4.  Fluorine and Halogens are always -1 if they are the most electronegative

5.  Hydrogen ALWAYS +1…except… à when with other metal (i.e. LiH vs. HCl)

6.  OXYGEN is usually -2 in compounds: ex) In H2O = ex) With Fluorine = 7.  The SUM of all Oxidation #s in compound =

ZERO 8.  The SUM of all Oxidation #s in polyatomic

ion must = CHARGE on the ion

OXIDATION #’S

+2 bc “F” more electronegative -2

What are the oxidation #s of the atoms in HNO3? We know…HNO3 à (1) H // (1) N // (3) O’s ¡ Hydrogen: oxidation # = ¡ Oxygen: each oxidation # = § SO total for Oxygen =

¡ Oxidation # of ENTIRE compound must equal (0), therefore: § (+1) + (-6) =

OXIDATION #’S

+1 (rule 5) -2 (rule 6)

-6 (from 3 total O's)

-5 …sooo

H = (+1) O = (-6) N = (+5)

What is the oxidation # of Chromium in the dichromate ion (Cr2O7

2-)? We know…Cr2O7 à (2) Cr // (7) O’s ¡ Oxygen: oxidation # =

¡ Sum of Oxidation # of ION must equal the ion’s CHARGE, therefore: § Cr2O7 ion has a charge of: § 7 O’s + 2Cr’s = (-14) + 2x = -2

2x = +12 x = +6 = Oxid. # of Cr

OXIDATION #’S

-2 (rule 6)

-2

O = (-14)

How do we know if a reaction is REDOX? ¡ Assign Oxidation #s (states) § if there is a change in oxidation # for a

particular type of atom, then it’s redox ¡ GIVE AWAY à if an uncombined element is

on one side of an equation and is combined in a compound on the other side = REDOX

double replacement reactions = NOT REDOX REDOX: 2Zn + 2HCl à ZnCl2 + H2 NOT rdx: NaCl + AgNO3 à AgCl + NaNO3

(double replacement=above)

REDOX REACTIONS

What’s being oxidized/reduced in a reaction? ¡ INCREASE in OXID. # = oxidation occurred ¡ DECREASE in OXID. # = reduction occurred

Ex) MnO2 + 4HCl à MnCl2 + Cl2 + 2H2O ¡ Identify all oxidation #’s à Cl: (-1) as reactant in HCl but as product = some (-1) in MnCl2 but in Cl2 = (0)

§ Because Cl changed from lower oxidation # (-1) to higher oxidation # (0) it’s been oxidized = lost e-

à Mn: (+4) as reactant & (+2) as product § Because Mn changed from higher oxidation # (+4)

to lower oxidation # (+2) it’s been reduced = gain e-

REDOX REACTIONS

…then WHAT’s an oxidizing/reducing agent? Ex) MnO2 + 4HCl à MnCl2 + Cl2 + 2H2O

We already precisely identified that: ¡ Mn is the atom undergoing REDUCTION (receiving

electrons from Cl-… Cl caused the reduction of the Mn+4 to Mn+2) § SO: Cl = the reducing “AGENT”

¡ Cl is the atom undergoing OXIDATION (losing electrons TO Mn…Mn+4 caused the oxidation of Cl-) § SO: Mn = the oxidixing “AGENT”

*the substance oxidized is the reducing agent *the substance reduced is the oxidizing agent

REDOX REACTIONS

¡ Chemical equations show the formulas of reactants and products

¡ HALF Reactions show oxidation/reduction portions of a redox reaction (e- gained/lost)

Ex) Fe3+(aq) + 3e- à Fe(s) ¡ reduction đ reaction: atom/ion gained one

or more electrons & oxidation # decreased Ex) Fe(s) à Fe3+(aq) + 3e-

¡ oxidation đ reaction: atom/ion LOST one or more electrons & oxidation # increased

*NET charge MUST be the same on both sides of the equation but DOES NOT have to be 0!*

HALF REACTIONS

Be Precise & determine the net charges: Ex) Fe3+(aq) + 3e- à Fe(s)

à Net charge/side: 0 Ex) Fe(s) à Fe3+(aq) + 3e-

à Net charge/side: 0 Ex) Sn4+(aq) + 2e- à Sn2+(aq)

à Net charge/side: 2+ Ex) Sn2+(aq) à Sn4+(aq) + 2e-

à Net charge/side: 2+

HALF REACTIONS

How to write a đ-reaction from an equation: 1.  Assign oxidation #’s to each element

Cu + AgNO3 à Cu(NO3)2 + Ag

2.  Write partial đ reactions to show change in oxidation state (include e-s & net charge) �

Oxidation: Cu à Cu2+ + 2e- net charge = 0

Reduction: Ag+ + e- à Ag net charge = 0

HALF REACTIONS

0 +1 +5 -2 +2 +5 -2 0

3.  Balance the # of electrons lost/gained (then the rest of the reaction)

Cu à Cu2+ + 2e- 2(Ag+ + e- à Ag)

so now… cancel out electrons on either side and add the two đ reactions…

Cu à Cu2+ + 2e- 2Ag+ + 2e- à 2Ag

2Ag+ + Cu à Cu2+ + 2Ag

HALF REACTIONS

¡ Redox reactions involve a flow of electrons ¡ Two Types of Electrochemical Cells: § voltaic cell: “Spontaneous” reaction producing

a flow of electrons § electrolytic cell: requires an electric current

to force a “non-spontaneous” reaction ¡ ELECTRODE à the site at which oxidation

or reduction occurs § Anode: where oxidation occurs § Cathode: where reduction occurs

“ANox & REDcat”

ELECTROCHEMICAL CELLS

¡ Salt Bridge: allows ions to migrate between half-cells, maintain electrical neutrality of the solutions § it makes a complete

circuit and allows charge to flow

§ chemical energy is spontaneously converted to electrical energy in this electrochemical cell

VOLTAIC CELLS

Zn(s)+ Cu2+(aq) à Cu(s) + Zn2+(aq) ¡ Zinc placed in Lead Nitrate

Solution ¡ Zn = oxidized & Cu ions = reduced

What’s going on? ¡ zinc strip in LEFT

beaker, Cu ions in RIGHT

¡ electric current produced because e-

s forced to flow through wire to complete the circuit

¡ E-s LOST (oxidation) from anode (Zn) flow through the wire to the cathode, which then GAINS those e-s, thereby continuing the cycle

VOLTAIC CELLS

Zn(s)+ Cu2+(aq) à Cu(s) + Zn2+(aq) ¡ Zinc placed in Lead Nitrate

Solution ¡ Zn = oxidized & Cu ions = reduced

Using your Reference Table J: activity series ¡ can be used to identify the anode and cathode in a

voltaic cell ¡ the higher metal on the chart of the 2 identified in

a cell is being OXIDIZED @ ANODE ¡ the lower metal on the chart of the 2 identified in

a cell is being REDUCED @ CATHODE ex) Zn0(s) + Pb2+(aq) à Zn2+(aq) + Pbo(s)

Identify the anode & cathode & direction of e- flow à Zn is HIGHER on table J = more reactive…soOo…

à will get oxidized and is the ANODE à Pb2+ ions will be reduced at the CATHODE à Electron flow: from anode (Zn) to cathode (Pb)

VOLTAIC CELLS

¡ When a power source (electricity) forces a reaction to occur = Electrolysis

¡ Ex) CAR Battery v. Alternator § Spontaneous reaction occurs in

battery to “start” the car § Once started, the alternator, in a

non-spontaneous reaction, “charges” the battery

ELECTROLYTIC CELLS

What’s going on: ¡ Anode = Silver Plate (left) & oxidation produces ions ¡ Ag ions travel through the solution to the cathode ¡ Cathode = reduces ions BACK to Ag atoms & adhere to

the metal being “Plated”

ELECTROLYTIC CELLS Tips: ¡ Like charges repel so ANODE is + ¡ CATHODE is - charge since

electrons migrate toward it ¡ External power source forces

electrons in wire to travel from anode to cathode

Voltaic v. Electrolytic: ¡ BOTH: redox, AnOX & RedCAT,

wire carries e-s from Anode à Cathode

¡ Voltaic: spontaneous, Anode = (-) & Cathode = (+) ¡ Electrolytic: non-spontaneous, Anode = (+) &

Cathode = (-)