Download Oxidation-Reduction (Redox) Reactions

Oxidation-Reduction (Redox)
Reactions and
Electrochemistry
Dr. Joshi
Graphics and examples taken from Chemistry, 3rd ed
by Olmsted and Williams.
Electrochemistry
• Study of interchange between chemical
and electrical energy
• Use or generation of electric current in
conjunction with chemical reaction
What is redox?
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Oxidation is the loss of electron(s).
Reduction is the gain of electron(s).
LEO says GER
The two processes occur together.
2 Mg (s) + O2 (g) --> 2 MgO (s)
Mg --> Mg2+ + 2 e(half-reaction)
(half-reaction)
O2 + 4 e- --> 2 O2How do these half reactions get put together?
Which species is oxidized?
Which is reduced?
Which is the oxidizing agent?
Which is the reducing agent?
One of these is a redox reaction, and the other is not.
How can you tell which is which?
You need to determine which species, if any, has lost electrons
and which, if any, has gained electrons.
We do this by assigning oxidation numbers.
Rules for assigning oxidation states (oxidation numbers):
1.
An atom in its elemental state has an oxid. state of zero. e.g. Na, H2, O2, Br2, Mg
2.
An atom in a monatomic ion has oxid. number identical to its charge. e.g. Na+ is +1, Al3+ is +3,
Cl– is –1 You already know what types of ions many of the elements tend to form!
3.
An atom in a polyatomic ion or a compound has the same oxid. # it would have if it were an
ion. e.g. MgCl2 Mg is +2, and Cl is –1.
4.
Oxygen is almost always –2. O takes precedence over halogens and H. (Exception: peroxide,
an O-O bond. You won’t run into this very much.)
5.
Halogens are –1. Exception: when bound to O. O takes precedence! Cl–O–Cl
Each Cl is
+1 b/c O has to be –2.
6.
H is typically +1 when it’s with nonmetals and –1 when it’s with metals.
Oxygen and metals take precedence.
H2O
O has to be –2, so each H is +1.
NaH
Na has to be +1 (alkali metal), so H is –1.
7.
Left side of periodic table (metals) cation-like and right side (nonmetals) anion-like.
8.
Sum of oxidation numbers equals overall charge.
Advice: Remember which elements take precedence, and make the others adjust.
In a compound or as an ion, alkali metals will always be +1.
In a compound or an ion, alkaline metals will always be +2.
(The other atoms in the compound have to adjust their oxidation states to give the correct overall
charge. e.g. NaH)
O is –2 (unless two Os bonded to each other)
(Everything else in the compound must adjust.)
Assign oxidation numbers (states) to determine which
is the redox reaction.
Which species is oxidized? Which is reduced?
Balancing Redox Reactions
Step 1: Break the unbalanced equation into half-reactions.
Step 2: Balance each half-reaction by following a-d.
a. Balance all elements except O and H
by adjusting coefficients.
b. Balance O by adding H2O to the appropriate side.
c. Balance H by adding the appropriate number of H+
ions to the side that needs H.
Add H2O to the other side if needed.
d. Balance net charge by adding electrons.
Step 3: Multiply the half-reactions by integers that will allow
for cancellation of electrons.
Step 4: Combine half-reactions, and simplify by combining
and canceling duplicated species.
Step 5 (Only if the solution is under basic conditions): Add
enough OH- (to both sides) to neutralize any H+ ions.
Simplify by canceling duplicate species (if needed).
Balance the following redox reaction (first under acidic/neutral
conditons, then under basic conditions).
Cu (s) + NO3— (aq) --> Cu2+ (aq) + NO (g)
Which metal is oxidized? Which is reduced?
Is this process spontaneous?
Oxidation occurs
at the anode.
Reduction occurs
at the cathode.
The anode and
cathode are
electrodes.
Schematic view of an arrangement that allows a sustained redox reaction
accompanied by an external flow of electrons.
Using Standard Reduction Potentials
Zn2+(aq) + 2 e- (wire)
Zn(s)
Cu2+(aq) + 2 e- (wire)
Eo = + 0.76 V
Cu(s) Eo = + 0.34 V
See table of Standard reduction potentials.
Change sign of E when reverse the half-reaction.
E does NOT change when half-rxn is multiplied!
Add up E values when add half rxns together.
DGo =
- nFE
o
n = # of mol of eF = Faraday’s constant = 96,485 C/mol ePositive cell potential gives negative DG.
So, positive E means the process is spontaneous.
Standard conditions: 1 M, 1 atm
Line Notation:
Pt(s)
IH
2
(g), H+ (aq)
II Fe
3+
(aq), Fe2+ (aq)
I Pt(s)
Passive Electrodes
Fe3+ + e- --> Fe2+
H2 + 2H2O --> 2H3O+ + 2e-
A galvanic cell runs spontaneously (E is positive).
An electrolytic cell does not run spontaneously (E is negative).
To run an electrolytic cell, electric current must be supplied.
Nernst Equation:
Cell Potentials Under Non-Standard Conditions
E = Eo – RT ln (Q)
nF
For a cell at equilibrium,
E=0
and
Q=K
log (K) = nEo
0.0591