Transcript slides

An informative exploration by
JC and Petey B.
Oxidation Numbers
All oxidation and reduction reactions involve the transfer of electrons
between substances.
Ag+ accepts electrons from
Cu and is reduced to Ag;
Ag+ is the oxidizing agent.
2 Ag+(aq) + Cu(s)  2 Ag(s) + Cu2+(aq)
Losing electrons means oxidation.
Gaining electrons means reduction.
Cu donates electrons to
Ag+ and is oxidized to Cu2+;
Cu is the reducing agent.
What are oxidation numbers?
The oxidation number of an atom in a
molecule is defined as the electric charge
an atom has.
Example: Al3+
– Al3+ has an oxidation number of +3.
But how do we do this with more
complicated molecules?!
Tips for determining oxidation numbers
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
the charge of the ion.
Fluorine is always -1 in compounds with other elements.
Cl, Br, and I are always -1 in compounds except when combined
with O or F.
The oxidation number of H is +1 and of O is -2.
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
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
with acid.
C2H5OH(aq) + Cr2O72-(aq)  CH3CO2H(aq) + Cr3+(aq)
First, identify what is being oxidized and reduced:
– Cr (+6+3); Cr2O72- is being reduced.
– C (-2 0); C2H5OH is being oxidized.
Find the two half reactions:
– C2H5OH  CH3CO2H
– Cr2O72-  Cr3+
Then, balance the half reactions for mass:
– C2H5OH + H2O  CH3CO2H + 4 H+
– 14 H+ + Cr2O72-  2 Cr3+ + 7 H2O
Now, balance the half reactions for charge:
– C2H5OH + H2O  CH3CO2H + 4 H+ + 4 e– 6 e- + 14 H+ + Cr2O72-  2 Cr3+ + 7 H2O
We then take the half reactions and multiply them by the appropriate factors
to make the number of electrons on each side equal:
– 3 [C2H5OH + H2O  CH3CO2H + 4 H+ + 4 e-]
– 2 [6 e- + 14 H+ + Cr2O72-  2 Cr3+ + 7 H2O]
Add the two balanced half reactions:
– 3 C2H5OH + 3 H2O + 12 e- + 28 H+ + 2 Cr2O72- 
–
3 CH3CO2H + 12 H+ + 12 e- + 4 Cr3+ + 14 H2O
Eliminate commons reactants and products:
– 3 C2H5OH(aq) + 16 H+(aq) + 2 Cr2O72-(aq)  CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l)
That was a pretty basic example. Here’s
an example that’s even more basic!
The “Basic” Concept
Given: SnO22-(aq)  SnO32-(aq)
First, note the change in oxidation number of Sn, from +2
to +4.
Separate into half reactions (this one is already done).
Balance for mass:
– Since the left side is deficient in oxyigen, add the
oxygen-rich OH-.
– For every two OH-’s we use, we need one H2O on the
opposite side.
– So: 2 OH- + SnO22-  SnO32- + H2O
Next, balance for charge:
– 2 OH- + SnO22-  SnO32- + H2O + 2 e-
Electrochemical Cells
Electrochemical cells are very closely
related to oxidation-reduction reactions
because the transfer of electrons results in
a potential difference.
The rest of this slide is empty.
– Move on to the next one.
Or else.
How it works
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Electrons flow through the wire from
the Zn electrode (anode) to the Cu
electrode (cathode).
A salt bridge provides a connection
between the half-cells for ion flow;
thus, SO42- ions flow from the copper
to the zinc compartment.
Electrons always flow from the anode
to the cathode.
–
(mnemonic: alphabetical order)
Oxidation takes place at the anode,
reduction at the cathode.
–
AnOx, RedCat
Ions flow through the salt bridge in the
opposite direction of the electrons.
An Electro-Demo
All About Potential
The standard potential Eo is a quantitative
measure of the tendency of the reactants
in their standard states to proceed to
products in their standard states.
Free energy is associated with the same
characteristics, and is defined as:
– ∆Gorxn= -nFEo
where n=number of moles of electrons transferred
in a balanced redox reaction and f is the faraday
constant, 9.65x104
Calculating Cell Potential
Given: 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=EoZn+EoNi Therefore, EoNi = 0.51-0.76= -0.25V
Ta da!
A note about using
the table of standard reduction potentials
All potentials listed are for reduction reactions; the sign
must be switched for oxidation.
All half reactions are reversible.
The more positive the value of the reduction potential,
the reaction as written is more likely to occur as a
reduction. Given two half reactions, the one with the
more positive Eo is the one that will occur as on
oxidation.
Changing the stoichiometric coefficients for a halfreaction does not change the value of Eo.
Non-standard conditions
E=Eo – (RT/nF)ln(Q)
– Q= Reaction quotient
Q=[products]/[reactants]
– F= Faraday constant
9.65x104 joules/(volts*mole)
– R= Gas constant
8.315 joules/(K*mole)
Mass  Current
Current I (amperes, A) =
– charge (coulombs, C) / time (seconds, s)
Example:
– 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 cathode.
What mass of silver is deposited?
Ag+(aq) + e-  Ag(s
– Calculate the charge passed in 15.0 minutes:
Q=I*t=(1.5A)(15.0 min)(60.0 sec/min)=1350 C
– Next, calculate the number of moles of electrons:
(1350 C) ((1 mol e-)/(9.65x104 C))=0.0140 mols e-
– Finally, calculate mass of silver deposited:
(0.0140 mols e-) ((1 mol Ag)/(1 mol e-)) ((107.9 g Ag)/(1 mol Ag))=
1.51 g Ag
Credits
Special thanks to:
The video camera
The book
The still camera
The video capture box
Delicious water (H2O(l))
And of course…
The Skipster himself
Our grades just dropped. A lot. Oh well.
The End
The End
– But to be continued???
No, probably not