An informative exploration An informative exploration by JC and Petey B.by JC and Petey B.

Oxidation NumbersOxidation NumbersAll oxidation and reduction reactions involve the transfer of electrons between substances.

2 Ag+(aq) + Cu(s) 2 Ag(s) + Cu2+

(aq)

Ag+ accepts electrons from Cu and is reduced to Ag; Ag+ is the oxidizing agent.

Cu donates electrons to Ag+ and is oxidized to Cu2+; Cu is the reducing agent.Losing electrons means oxidation.

Gaining electrons means reduction.

What What areare oxidation numbers? oxidation numbers?

The oxidation number of an atom in a The oxidation number of an atom in a molecule is defined as the electric charge molecule is defined as the electric charge an atom has.an atom has.

Example: AlExample: Al3+3+

– AlAl3+3+ has an oxidation number of +3. has an oxidation number of +3.

But how do we do this with more But how do we do this with more complicated molecules?!complicated molecules?!

Tips for determining oxidation numbersTips for determining oxidation numbersEach atom in a pure element has an oxidation number of zero.Each atom in a pure element has an oxidation number of zero.

For ions consisting of a single atom the oxidation number is equal to For ions consisting of a single atom the oxidation number is equal to the charge of the ion.the charge of the ion.

Fluorine is always -1 in compounds with other elements.Fluorine is always -1 in compounds with other elements.

Cl, Br, and I are always -1 in compounds except when combined Cl, Br, and I are always -1 in compounds except when combined with O or F.with O or F.

The oxidation number of H is +1 and of O is -2.The oxidation number of H is +1 and of O is -2.

The algebraic sum of the oxidation number in a neutral compound The algebraic sum of the oxidation number in a neutral compound must be zero; in an ion, the sum must be equal to the overall ion must be zero; in an ion, the sum must be equal to the overall ion charge.charge.

Example: Cr2O72-

First, recognize that the net charge must be -2.

Then, assign an oxidation number of -2 to the O’s.

(-2)*7 + (x)*2 = -2

Therefore, x = +6.

Balancing Redox Reactions: An example on Balancing Redox Reactions: An example on with acid.with acid.

CC22HH55OHOH(aq)(aq) + Cr + Cr22OO772-2-

(aq)(aq) CH CH33COCO22HH(aq)(aq) + Cr + Cr3+3+(aq)(aq)

First, identify what is being oxidized and reduced:First, identify what is being oxidized and reduced:– Cr (+6Cr (+6+3); Cr+3); Cr22OO77

2-2- is being reduced. is being reduced.

– C (-2 C (-2 0); C0); C22HH55OH is being oxidized.OH is being oxidized.

Find the two half reactions:Find the two half reactions:– CC22HH55OH OH CH CH33COCO22HH

– CrCr22OO772-2- Cr Cr3+3+

Then, balance the half reactions for mass:Then, balance the half reactions for mass:– CC22HH55OH + HOH + H22O O CH CH33COCO22H + 4 HH + 4 H++

– 14 H14 H++ + Cr + Cr22OO772-2- 2 Cr 2 Cr3+3+ + 7 H + 7 H22OO

Now, balance the half reactions for charge:Now, balance the half reactions for charge:– CC22HH55OH + HOH + H22O O CH CH33COCO22H + 4 HH + 4 H++ + 4 e + 4 e--

– 6 e6 e-- + 14 H + 14 H++ + Cr + Cr22OO772-2- 2 Cr 2 Cr3+3+ + 7 H + 7 H22OO

We then take the half reactions and multiply them by the appropriate factors We then take the half reactions and multiply them by the appropriate factors to make the number of electrons on each side equal:to make the number of electrons on each side equal:

– 3 [C3 [C22HH55OH + HOH + H22O O CH CH33COCO22H + 4 HH + 4 H++ + 4 e + 4 e--]]

– 2 [6 e2 [6 e-- + 14 H + 14 H++ + Cr + Cr22OO772-2- 2 Cr 2 Cr3+3+ + 7 H + 7 H22O]O]

Add the two balanced half reactions:Add the two balanced half reactions:– 3 C3 C22HH55OH + 3 HOH + 3 H22O + 12 eO + 12 e-- + 28 H + 28 H++ + 2 Cr + 2 Cr22OO77

2-2-

– 3 CH3 CH33COCO22H + 12 HH + 12 H++ + 12 e + 12 e-- + 4 Cr + 4 Cr3+3+ + 14 H + 14 H22OO

Eliminate commons reactants and products:Eliminate commons reactants and products:– 3 C3 C22HH55OHOH(aq)(aq) + 16 H + 16 H++

(aq)(aq) + 2 Cr + 2 Cr22OO772-2-

(aq)(aq) CH CH33COCO22HH(aq)(aq) + 4 Cr + 4 Cr3+3+(aq)(aq) + 11 H + 11 H22OO(l)(l)

That was a pretty basic example. Here’s an example that’s even more basic!

The “Basic” ConceptThe “Basic” ConceptGiven: SnOGiven: SnO22

2-2-(aq)(aq) SnO SnO33

2-2-(aq)(aq)

First, note the change in oxidation number of Sn, from +2 to +4.First, note the change in oxidation number of Sn, from +2 to +4.

Separate into half reactions (this one is already done).Separate into half reactions (this one is already done).

Balance for mass:Balance for mass:– Since the left side is deficient in oxyigen, add the oxygen-Since the left side is deficient in oxyigen, add the oxygen-

rich OHrich OH--..

– For every two OHFor every two OH--’s we use, we need one H’s we use, we need one H22O on the O on the

opposite side.opposite side.

– So: 2 OHSo: 2 OH-- + SnO + SnO222-2- SnO SnO33

2- 2- + H + H22OO

Next, balance for charge:Next, balance for charge:

– 2 OH2 OH-- + SnO + SnO222-2- SnO SnO33

2- 2- + H + H22O + 2 eO + 2 e--

Electrochemical CellsElectrochemical Cells

Electrochemical cells are very closely Electrochemical cells are very closely related to oxidation-reduction reactions related to oxidation-reduction reactions because the transfer of electrons results in because the transfer of electrons results in a potential difference.a potential difference.

The rest of this slide is empty.The rest of this slide is empty.– Move on to the next one.Move on to the next one.

Or else.Or else.

How it worksHow it works

ZnZn(s)(s) + Cu + Cu2+2+(aq) (aq) ZnZn2+2+

(aq) (aq) + Cu+ Cu(s)(s)

Electrons flow through the wire from Electrons flow through the wire from the Zn electrode (anode) to the Cu the Zn electrode (anode) to the Cu electrode (cathode). electrode (cathode).

A salt bridge provides a connection A salt bridge provides a connection between the half-cells for ion flow; between the half-cells for ion flow; thus, SOthus, SO44

2-2- ions flow from the copper to ions flow from the copper to

the zinc compartment.the zinc compartment.

Electrons always flow from the anode Electrons always flow from the anode to the cathode.to the cathode.

– (mnemonic: alphabetical order)(mnemonic: alphabetical order)

Oxidation takes place at the anode, Oxidation takes place at the anode, reduction at the cathode.reduction at the cathode.

– AnOx, RedCatAnOx, RedCat

Ions flow through the salt bridge in the Ions flow through the salt bridge in the opposite direction of the electrons.opposite direction of the electrons.

An Electro-DemoAn Electro-Demo

All About PotentialAll About Potential

The standard potential EThe standard potential Eoo is a quantitative is a quantitative measure of the tendency of the reactants measure of the tendency of the reactants in their standard states to proceed to in their standard states to proceed to products in their standard states. products in their standard states. Free energy is associated with the same Free energy is associated with the same characteristics, and is defined as:characteristics, and is defined as:– ∆∆GGoo

rxnrxn= -nFE= -nFEo o

where n=number of moles of electrons transferred where n=number of moles of electrons transferred in a balanced redox reaction and f is the faraday in a balanced redox reaction and f is the faraday constant, 9.65x10constant, 9.65x1044

Calculating Cell PotentialCalculating Cell PotentialGiven: the cell illustrated has a potential of: EO=+0.51 V at 25 oC.

The net ionic equation is:

Zn(s) + Ni2+(aq, 1M) Zn2+

(aq, 1M) + Ni(s)

What is the value of Eo for the half-cell :

Ni2+(aq) + 2e- Ni(s) ?

Solution: For the anode, Zn, we know the potential is +0.76 V from the table of standard reduction potentials. Note, we had to change the sign from the table because our reaction is: Zn(s) Zn2+

(aq) + 2e-.

Since the net reaction is the sum of the half reactions:

Eonet=Eo

Zn+EoNi Therefore, Eo

Ni = 0.51-0.76= -0.25V

Ta da!

A note about using A note about using the table of standard reduction potentialsthe table of standard reduction potentials

All potentials listed are for reduction reactions; the sign All potentials listed are for reduction reactions; the sign must be switched for oxidation.must be switched for oxidation.

All half reactions are reversible.All half reactions are reversible.

The more positive the value of the reduction potential, The more positive the value of the reduction potential, the reaction as written is more likely to occur as a the reaction as written is more likely to occur as a reduction. Given two half reactions, the one with the reduction. Given two half reactions, the one with the more positive Emore positive Eoo is the one that will occur as on is the one that will occur as on oxidation. oxidation.

Changing the stoichiometric coefficients for a half-Changing the stoichiometric coefficients for a half-reaction does not change the value of Ereaction does not change the value of Eoo..

Non-standard conditionsNon-standard conditions

E=EE=Eoo – (RT/nF)ln(Q) – (RT/nF)ln(Q)– Q= Reaction quotientQ= Reaction quotient

Q=[products]/[reactants]Q=[products]/[reactants]

– F= Faraday constantF= Faraday constant9.65x109.65x104 4 joules/(volts*mole)joules/(volts*mole)

– R= Gas constantR= Gas constant8.315 joules/(K*mole)8.315 joules/(K*mole)

Mass Mass Current Current

Current I (amperes, A) = Current I (amperes, A) = – charge (coulombs, C) / time (seconds, s)charge (coulombs, C) / time (seconds, s)

Example:Example:– A current of 1.50 A is passed through a solution containing silver ions A current of 1.50 A is passed through a solution containing silver ions

for 15.0 minutes. The voltage is such that silver is deposited at the for 15.0 minutes. The voltage is such that silver is deposited at the cathode. What mass of silver is deposited?cathode. What mass of silver is deposited?

AgAg++(aq)(aq) + e + e-- Ag Ag(s(s

– Calculate the charge passed in 15.0 minutes:Calculate the charge passed in 15.0 minutes:Q=I*t=(1.5A)(15.0 min)(60.0 sec/min)=1350 CQ=I*t=(1.5A)(15.0 min)(60.0 sec/min)=1350 C

– Next, calculate the number of moles of electrons:Next, calculate the number of moles of electrons:(1350 C) ((1 mol e(1350 C) ((1 mol e--)/(9.65x10)/(9.65x1044 C))=0.0140 mols e C))=0.0140 mols e--

– Finally, calculate mass of silver deposited:Finally, calculate mass of silver deposited:(0.0140 mols e(0.0140 mols e--) ((1 mol Ag)/(1 mol e) ((1 mol Ag)/(1 mol e--)) ((107.9 g Ag)/(1 mol Ag))=)) ((107.9 g Ag)/(1 mol Ag))=

1.51 g Ag1.51 g Ag

CreditsCredits

Special thanks to:Special thanks to:

The video cameraThe video camera

The bookThe book

The still cameraThe still camera

The video capture boxThe video capture box

Delicious water (HDelicious water (H22OO(l)(l)))

And of course…And of course…

The Skipster himselfThe Skipster himself

Our grades just dropped. A lot. Oh well.

The EndThe End

The EndThe End– But to be continued???But to be continued???

No, probably notNo, probably not