Transcript Document
Chuga
Chuga
Chuga
Chuga Choo
Adam Rosenbloom and
Olga Lozovskaya
Choo!
This one’s for you, Mr. Hinton
Oxidation Numbers
A number assigned to each element in a compound in order to keep
track of the electrons during a reaction
The oxidation number of pure elements’ atoms is zero (Br2)
For monatomic ions, the oxidation number equals the ion’s charge
(K+)
Fluorine is always -1 in compounds with other elements.
Cl, Br, and I are -1 unless they are with oxygen or fluorine.
H is almost always +1 and O is almost always -2.
The sum of the oxidation numbers in a neutral compound must be
zero; in a polyatomic ion the sum must equal the overall charge of the
ion.
Redox Reactions
In an oxidation-reduction reaction, there is a transfer of electrons from one
species to another.
The reactant that is oxidized loses an electron, and its oxidation number
increases.
The reactant that is reduced gains an electron, and its oxidation number
decreases.
Example:
A + B An+ + BnA is being oxidized and since it donates an electron to B, it is also the
reducing agent.
B is being reduced and since it accepts an electron from A, it is the oxidizing
agent.
Balancing Redox Reactions
(Acidic Solution)
Ag+ (aq) +HCHO(aq)Ag(s) +HCO2H(aq)
Step 1: Divide into two half reactions
Ag+ (aq) Ag(s)
HCHO(aq) HCO2H(aq)
Step 2: Balance elements other than H and O.
Step 3: Balance O by adding H2O.
HCHO(aq) + H2O HCO2H(aq)
Step 4: Balance H by adding H+.
HCHO(aq) + H2O HCO2H(aq) + 2H+
Step 5: Balance charge by adding electrons:
Ag+ (aq) + e- Ag(s)
HCHO(aq) + H2O HCO2H(aq) + 2H+ + 2eStep 6: Make electrons cancel so you can add the half-reactions.
2(Ag+ (aq) + e- Ag(s) )
Step 7: Combine the two half-reactions:
+
+
2Ag + HCHO(aq) + H2O HCO2H(aq) + 2H + Ag(s)
Balancing Redox Reactions
(Basic Solution)
This is the same process as with an acidic solution, except that Steps 3
and 4 are different:
Step 1: Divide into two half reactions
Step 2: Balance elements other than H and O.
Step 3: Balance O by adding OH-.
Step 4: Balance H by adding H2O.
Step 5: Balance charge by adding electrons.
Step 6: Make electrons cancel so you can add the half-reactions.
Step 7: Combine the two half-reactions.
Electrochemical Cells
A current is generated by immersing
metals with different reduction potentials
into solutions of their ions connected by a
salt bridge.
An electrochemical cell produces electric
current due to the transfer of electrons
from the anode to the cathode.
At the negative electron, the cathode,
reduction takes place and electrons are
deposited. The cathode gains mass.
At the positive electron, the anode,
oxidation takes places and electrons are
given off. The anode loses mass.
Example of a Wet Cell
The reduction
potential of Zinc (II) is
–0.763 V. The
reduction potential of
Ni is –0.25 V. Since
the Ni has the higher
reduction potential, it
experiences reduction.
The half-reaction for
zinc is then flipped for
it to be oxidized.
To get E0 (the standard
electric potential),
subtract the substance
oxidized from the one
that is reduced. (-0.25-0.763 = 0.513 V)
TO BE CONTINUED…
E0 continued…
The equation for the relationship between E0 and Gibbs Free
Energy is:
ΔG0rxn = -nFE0
n is the number of moles of electrons transferred
F is the is the charge of one mole of electrons, which is the Faraday
Constant (9.6485309 x 104 Joules / V / mol)\
Also, don’t forget that even if you need to multiply a half-reaction by
some factor to balance the number of electrons transferred, DO
NOT (upon penalty of death) multiply the E0 too.
Using the Faraday Constant for MassCurrent Relationships
A current of 2.3 amps is passed through a solution
containing potassium ions for 17 minutes. The
voltage is such that potassium is deposited at the
cathode. What mass, in grams, of potassium is
deposited at the cathode?
Step 1: Calculate the charge (number of Coulombs)
passed in 17 minutes.
Step 2: Calculate the number of moles of electrons.
Charge= current * time
Charge=(2.3A)(17 min)(60 sec/min)=2346 C
2346 C (1 mol e- / 9.65 *104) = .0243 mol e-
Step 3: Calculate the number of moles of potassium
and then the mass of potassium deposited.
(.0243 mol e- )(1 mol K/ 1 mol e-)(39.1 g/1 mol K) = .95
g of Potassium
I have no
idea what’s
going on
right now.
Cells at Non-Standard
Conditions
Nernst Equation- This equation can
help you correct E0 to be a
proper value under nonstandard conditions.
E = E0 – (RT / nF) ln (Q), where Q is
the reaction quotient and R is 8.314510
J / K / mol
I bet you
forgot
about
me…
When E = 0, that means the
reaction is at equilibrium, so
E0 = RT / nF ln (Keq)
Thank you Marc Potempa and Matt
Meshulam for this lovely information.
Thanks, electrochemistry!
Good-bye!