Transcript Electrochemistry

Electrochemistry
By: Matt Meshulam and
Marc Potempa
We Basically Have No Idea What
We Are Doing Here
Redox Reactions
An oxidation-reduction reaction is the
transfer of electrons between ions.
Say you have the reactants X and Y
X + Y  Xn+ + Yn X is oxidized, as it loses an electron
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
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Oxidation Half-Reaction: X  Xn+ + e-
Y is reduced, as it gains an electron

Reduction Half-Reaction: Y + e-  Yn-
Were you expecting something amusing?
Determining Oxidation
Numbers
1) Overall charge of a molecule is the
sum of its atoms’ oxidation numbers.
2) F, Cl, Br, I are –1. H is +1. O is –2.
3) H, O, and F take priority, in that order.
4) Pure element (H2, etc) has oxidation number of
zero.
Example: H2SO4 is neutral. 2(+1) for H, 4(-2) for O, therefore S is 1(+6).
Didn’t see a
catchy title
coming did ya?
But if you did,
couple other steps) touché. Touché.
Just add water,
(and a few other things, and combine with a
In order to balance a Redox Reaction:
Example Eq. – Cr2O72- + Fe2+  2Cr3+ + Fe3+


1) Split up the equation into half-reactions:
2)
Cr2O72-  2Cr3+
Fe2+  Fe3+
Add H+ and H2O (where necessary) and balance using e-
Cr2O72- + 14H+ + 6e-  2Cr3+ + 7H2O
(Fe2+  Fe3+ + e-) x 6 [to balance the electrons between ½ Rx’s]
 3) Combine into overall reaction.
Cr2O72- + Fe2+ + 14H+ 2Cr3+ + 6Fe3+ + 7H2O
What?! More Balancing?
Cr2O72- + Fe2+ + 14H+ 2Cr3+ + 6Fe3+ + 7H2O
To Balance in a Basic
+ 14OH+ 14OHSolution, begin by balancing
2- + Fe2+ + 14H O  2Cr3+ + 6Fe3+ + 7H O
Cr
O
2
7
2
2
as you would in an acidic
+ 14OH
+
solution using H
Next, add OH- to the H+, and
2- + Fe2+ + 7H O  2Cr3+ + 6Fe3+ +
Cr
O
,
2
7
2
form H2O (balancing both
14OH
sides of the equation with
the same amount of OH-)
As a last check, cancel out
the H2O to simplify the
equation.
Electrochemical Cells
Uh… we got nothing. Sorry.
An EC is a device that produces an electric current
as a result of an electron transfer reaction.

In other words, electricity from the process of oxidizing one
substance, and reducing another substance.
In all EC’s oxidation occurs at the anode, and
reduction occurs at the cathode.
A typical setup will include:


Two metal bars in either a sulfate or nitrate solution.
A salt bridge (which allows the anions to pass toward the site
of oxidation to restore charge).
EC-sample
The Zinc Electrode is:
An Anode
Place of Oxidation
Losing Mass
Source of Electrons that
transfer through the wire
The Copper Electrode is:
A Cathode
Place of Reduction
Gaining Mass
Receiving Electrons from
Anode
Some Stuff You’ll Want to
Know about Cell Potential
Standard Electrode Potential (E0)- The potential of an
electrochemical cell measured under standard conditions. This
value shows how much energy must be released or gained for
the reaction to occur.

Use Electrode Potential to determine the cell potential.


Electrode Potential is written for REDUCTION half-reactions. The
value for these half-reactions will generally be listed in a table.


E0cell = E0red – E0ox (Measured in Voltage [V])
To find the value for Oxidation half-reactions Just flip the sign on the
value, as oxidation is just the opposite of reduction.
Important note! Should the reaction need a half-reaction to multiply
by some factor, the E0 does NOT multiply as well.
Alright! Another Example!
Bet you love that.
Determine the spontaneous cell reaction and the cell potential of
a cell that has these two half reactions
Al3++3e-Al(s)
EoAl3+= -1.66V
Cu2++2e-Cu(s)
EoCu2+= 0.34V
First Determine which species will oxidize, and which will reduce.
The Oxidized substance will always have the lesser potential.
(ie. the E0cell should be positive)
Since we now know that Aluminum will be oxidized, we can use the
equation to find the overall voltage.
E0cell = .34 – (-1.66) = 2.00 V
Yes, Mr. Hinton, the question is ripped straight from last
year’s PowerPoint.
Some More Stuff to Know that
Involves E0
The relationship between E0 and Gibbs Free NRG is
determined by:
(energy, come on Professor Michael Chuy Bilow used this one)
ΔG0+rxn = -nFE0
where n is the number of moles of electrons transferred
F is the Faraday Constant (9.6485309 x 104 Joules / V / mol)
the charge of a mole of electrons
Nernst. Just Nernst.
Nernst Equation- This equation can help you correct E0
to be a proper value under non-standard conditions.
E = E0 – (RT / nF) ln (Q), where Q is the reaction quotient
and R is 8.314510 J / K / mol
When E = 0, that means the reaction is at equilibrium, so
E0 = RT / nF ln (Keq)
That was it…what are you still looking here for?