Redox

 Redox involves two simultaneous reactions ◦ An oxidation and a reduction  Oxidation involves a loss of electrons  Reduction involves a gain of electrons  LEO the lion says GER

 Seeing how in each Redox reaction, there is said to be an oxidation half-reaction and a reduction half-reaction  One molecule will lose electrons (the oxidation) and the electrons will join another atom (reduction)

 Oxidation half reaction ◦ Fe (s) -> Fe 2+ (aq) + 2e -  Iron is being oxidized to form the ferrous ion  Reduction half-reaction ◦ Au 3+ (aq) + 3e - -> Au (s)  The gold ion has been reduced to its ground state

AgentElectron exchangeHalf-reactionThe atom is Reducing AgentLoses electronsOxidationOxidized Oxidizing AgentGains electronsReductionReduced

 Oxidation half-reaction Zn (s) ->Zn +2 +2e - Cu 2+ +2e - ->Cu (s) ______________________________________________ Cu 2+ +Zn (s) ->Zn 2+ +Cu (s) By eliminating the common terms on either side, you are left with a simplified final equation

 If there is a difference in the number of electrons in the equations, you must use stoichiometrics!  Fe (s) -> Fe 2+ (aq) + 2e - x3  Au 3+ (aq) + 3e - -> Au (s) x2  3 Fe (s) + 2 Au 3+ (aq) -> 3 Fe 2+ (aq) + 2 Au (s)

 The reaction of a piece of magnesium (Mg) in hydrochloric acid (HCl) results in the formation of magnesium dichloride (MgCl 2 ). The release of hydrogen (H 2 ) can also be observed. During this reaction, metallic magnesium is oxidized into aqueous Mg 2+ ions, while aqueous H + ions of the acidic solution is reduced to hydrogen gas.  a) What are the half-reactions in this reaction?  b) What is the general equation for oxidation- reduction?  c) Find the oxidizing agent and the reducing agent

 a) Magnesium oxidized ◦ Loses electrons  Oxidation reaction ◦ Mg (s) -> Mg 2+ (aq) + 2e -  Hydrogen reduced ◦ Gains electrons  Reduction reaction ◦ H + (aq) + 1e - -> 1/2 H 2 (g)

 b)Add the two half reactions together  Mg (s) ->Mg 2+ (aq) +2e -  2x(H + (aq) + 1e - ->½ H 2 (g) )  2H + (aq) +2e - ->H 2 (g) ______________________________________________  Mg (s) + 2H + (aq) ->Mg 2+ (aq) + H 2 (g)

 c) The oxidizing agent is the H + (aq) because it gains electrons  The reducing agent is the Mg (s) because it loses electons  Remember LEO the lion says GER

 Do # 1, 2, 4, 5 on the worksheet  We will go over it next class  Have Fun!

Oxidation State

 The oxidation number, also called the oxidation state, indicates the number of electrons an element has lost or gained, in relation to its ground state, during a redox reaction.  All elements in their ground state have an oxidation number of 0  They are considered to be atoms, due to them not having lost any electrons

 When atoms are involved in redox reactions, their oxidation numbers vary  Oxidation numbers increase with an oxidation due to a loss in electrons  Oxidation numbers decrease with a reduction due to a gain in electrons

 To determine the oxidation number of an atom, we must determine whether it is part of an ionic or a covalent compound  Ionic compound ◦ A bond between a metal and a non-metal which share electrons  The oxidation number of each ion is equal to its charge

 Example  Calcium chloride (CaCl 2 ) is composed of one Ca 2+ ion and two Cl - ions  To distinguish between an ion’s charge and its oxidation number, the convention is different  A charge is written as 2+ while an oxidation number is written as +2

 Find the oxidation # for the following atoms AtomChargeOxidation # Na1++1 Sr2++2 Ra2++2 K1++1 Li1++1 Mg2++2 Rb1++1

Substances ChargeOxidation # Elements in ground state (Li, Mg, Al, Fe, etc.)00 Molecules of elements (H 2, O 2, Cl 2, N 2, S 8, etc.)00 Ions of alkali metals (Li +, Na +, K +, etc.)1++1 Ions of alkaline earth metals (Ca 2+, Mg 2+, Be 2+, etc.) 2++2

 When an atom is part of a molecule or polyatomic ion, convention determines its oxidation number by assigning each pair to the more electronegative atom in the bond, that is, the atom that is more likely to attract electrons to fill its outermost shell.  To determine the oxidation number of a molecule, the molecule can be represented with Lewis notation

 If you take a water molecule, you can see that the electrons are shared between the oxygen and the hydrogen  Since oxygen normally has 6 electrons ◦ It gets 2 additional ones from hydrogen ◦ Oxidation number of -2  Hydrogen atom has lost its electron ◦ Oxidation number of +1

 In covalent molecules, the charge is not always the same as the oxidation number  There is no O 2- or H + in water  In covalent compounds, the electrons are not given to the other atom, but rather are shared between the two  Let’s try this again...

 What are the oxidation numbers for the atoms in an ammonia (NH 3 ) molecule?

 1- Since there are no metallic atoms, we know that ammonia is a covalent compound  2- Using the periodic table, we can determine which molecule is more electronegative. In this case, it is Nitrogen. The electrons in the bond are assigned to it. Nitrogen needs 3 electrons to reach its ground state. The oxidation number of Nitrogen is -3

 3- As for Hydrogen, each hydrogen ion has lost its one electron, which gives it an oxidation number of +1  Ans: The oxidation number of nitrogen is -3 and hydrogen is +1

 The same element can have a different oxidation number depending in which molecule it is found in.  There are a few non-metals that almost always have the same oxidation number ◦ Oxygen is always -2, except in peroxide (-1)  Hydrogen is also almost always the same, +1

 What is the oxidation number for the manganese atom in the permanganate ion (MnO 4 - ), a very strong oxidizing agent? The permanganate ion’s negative charge indicates that the net charge is 1-. Furthermore, since the oxidation number of oxygen is -2 and there are 4 oxygen atoms in this molecule. How would you be able to figure out the oxidation number for the manganese?  Hint: math!

 Mn+4 O=-1  X+4 (-2)=-1  X+-8=-1  X=7

 What are the oxidation numbers of all of the atoms in the following redox reaction Fe 2 O 3 (s) + 2 Al (s) -> Al 2 O 3 (s) + 2 Fe (l)

 Seeing how Fe (s) and Al (s) are both solid, their oxidation numbers are 0  For Fe 2 O 3 and Al 2 O 3, oxygen has an oxidation number of -2 ◦ There are three of them, for a total of -6  The iron and aluminum need an oxidation number of +3 to balance everything out

 Metals can be classified by their reduction power  Spontaneous reactions  For spontaneous reactions to occur, the stronger reducing agent must be in the solid state and the weaker one must be in its ionic form

ReactionStronger Reducing AgentWeaker Reducing Agent SpontaneousSolidAqueous solution NoneAqueous solutionSolid

 You have Aluminum (Al) and Copper (Cu)  Which would have to be solid and which would have to be aqueous?  Same situation, but with Gold (Au) and Copper (Cu)  Which would have to be solid and which would have to be aqueous?

 For the Aluminum (Al) and Copper (Cu) reaction, the Al would need to be solid and the Cu in solution  For the Gold (Au) and Copper (Cu) reaction, the Cu would have to be solid and the Au in solution

 An electrochemical cell is a device that can spontaneously generate an electrical current ◦ i.e. A battery

 The electrochemical cell is composed of two electrodes, also called half-cells.  They are called half-cells due to the half reactions taking place in the cell  The two electrodes are joined by a wire and are connected by a salt bridge ◦ A tube of salt water plugged with a porous membrane

 The classic example of an electrochemical cell is a piece of zinc immersed in a solution of zinc sulfate (ZnSO 4 ) and a piece of copper in a solution of copper sulfate (CuSO 4 )  Since zinc is a better reducing agent than copper, it gives up its electrons to the copper  Positive terminal is called the anode and the negative is the cathode  Zinc is oxidized and the copper is reduced

 The concentration of each half-cell’s ions varies ◦ The Zn 2+ ions increase ◦ The Cu 2+ ions decrease  The ions in the salt bridge move towards their respective poles to maintain the number of positive and negative ions ◦ Positive towards the Cathode ◦ Negative towards the Anode

 During a redox reaction, one of the metals has greater potential energy than the other because it has a greater reducing power.  Without this potential difference between two metals of different types, no chemical reaction would take place  To calculate the potential, we need a reference electrode  In this case, we use the hydrogen electrode

 2 H e -  H 2  This reaction is considered to have a redox potential of 0.00 V  E O = 0.00V  Where E O is redox potential

 To measure the EO, calculate the potential difference, create an electrochemical cell using the Hydrogen and the electrode which you want to test  The voltage which is read on the voltmeter will be the redox potential of the unknown

 With this table, we can compare the oxidation or reducing power of substances  And, we can calculate the cell potential  That’s about it for you guys...

 The cell potential is equal to the sum of the oxidation potential and the reduction potential of the cell  Calculating the potential of an electrochemical cell made up of two different electrodes can be done without having to use a reference electrode

 First, determine the stronger reducing agent  If we use Silver (Ag) and Magnesium (Mg) as an example, the magnesium is stronger since it is found below silver in the table of redox potentials  This makes it the electrode which will lose electrons and undergo oxidation

 To find the cell potential, we must reverse the reduction. In this case, the Mg reaction will be the reduction  (Ag + + e - -> Ag (s) ) x2OxidationE O =0.80V  Mg (s) -> Mg e-ReductionE O =2.37V

 Cell potential is calculate using the following equation  E O cell = E O oxidation + E O reduction  E O cell = 0.80V V  E O cell = 3.17V

 The electrochemical cell can also be called a fuel cell  If the EO is above 0, then the reaction is said to occur spontaneously.  If the EO is below 0, then the fuel cell is said to not be spontaneous

 A fuel cell can be represented in this simplified way  OxidationReduction  Mg|Mg2+||Ag+|Ag  According to convention, the oxidation reaction is shown on the left