Unit 5: Electrochemistry
The oxidation number of an atom in a
molecule is defined as the electric charge an
atom has.
Balancing Redox Reactions: An example
with acid.
First, identify what is being oxidized and
reduced.
Find the two half reactions.
Then, balance the half reactions for mass.
Now, balance the half reactions for charge.
We then take the half reactions and multiply
them by the appropriate factors to make the
number of electrons on each side equal:
Add the two balanced half reactions:
Eliminate commons reactants and products:
Electrochemical cells are very closely
related to oxidation-reduction reactions
because the transfer of electrons results in a
potential difference.
Electrons flow through the wire from the Zn
electrode (anode) to the Cu electrode
(cathode).
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.
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
All potentials listed are for reduction
reactions; the sign must be switched for
oxidation.
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 half-reaction 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
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


by the non-Italian Stallion Peter Bartoszek
and the ever….present Jason Crystal