Redox reactions - Cardinal Newman
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Transcript Redox reactions - Cardinal Newman
Redox reactions
involve transfer of e-s
Mg0(s) + Cu2+(aq)
Cu0(s) + Mg2+(aq)
electrons transferred from Mg to Cu2+
Mg loses e-s and Cu2+ gains them
Mg is oxidized, loses e-s
Cu2+ is reduced, gains e-s
Redox reactions
involve transfer of e
Key terms
oxidation = loss of e-s (oxidation # goes up)
reduction = gain of e-s (oxidation # goes down)
oxidizing agent = e- acceptor (subst red)
reducing agent = e- donor (subst oxid)
Types of Electrochem Rx
Direct – oxidizing and reducing agents come
into contact with each other
Produce heat, not electrical energy
Indirect – oxidizing and reducing agents in
separate compartments, connected by
external wire that e-s pass thru
Each compartment called ½ cell
oxidation ½ cell
reduction ½ cell
½ cells are connected by salt bridge
Types Electrochemical cells
Galvanic or Voltaic cell
Apparatus that allows a redox rx to occur
Product favored rxs push e-s thru wire
thereby prod an electric potential (voltage)
Electrolytic cell
Reactant favored rx can be forced to occur
by using an electrical current to force a
nonspontaneous rx (electrolysis)
Parts of Electrochemical Cell
Oxidation ½ cell
Anode (-)
site of oxidation
e-s flow from it
Cathode (+)
site of reduction
e-s flow to it
Voltmeter
Reduction ½ cell
Measures cell potential (pressure)
Salt bridge
Allows ions to migrate
Parts of galvanic cell
1. Given the following data for Hg
Hg(l)
S˚ = 76.0 J/mol·K
Hg(g)
S˚ = 175.0 J/mol·K
calculate the normal boiling point for Hg:
2. For the reaction:
2H2O(l) + 2Br-(aq)
Hg(l)
Hf˚ = 0
KJ/mol
Hf˚ = 60.77 KJ/mol
Hg(g)
H2(g) + Br2(l) + 2OH-(aq)
a. calc G˚ at 25˚C.
b. calc G when pH2 = 0.100 atm, [Br-} = 0.750 M, and the pH of the solution is 9.73.
Cell potential = E˚
amount of push with which the e-s are
sent through the external circuit.
e-s are pushed through a galvanic cell by
electromotive force (emf)
the diff in electrical potential energies
(Ep) of the metals that make up the
anode and cathode determine the
potential (voltage), or electrical
pressure, of the cell
Units of Electricity
Coulomb = unit of electrical charge
Ampere = unit of electrical flow (current)
1 e- = 1.602 x 10-19 C
Amp = 1 C/sec
Joule = unit of energy
1J=1Cx1V
Energy = charge x potential (push)
Faraday = F = 96,480 C/mol e-
Determination of E˚
E = electrical potential under standard
conditions (1 atm, 1 molar)
E˚cell = E˚oxid + E˚red
E˚oxid = -(E˚red)
(+) value for E˚cell = rx is NOT spontaneous
(-) value for E˚cell = rx is spontaneous
Units = volts
1 volt = 1 J/1 C
Calculation of cell potential
Zn0(s) + Cu2+(aq)
Zn2+(aq) + Cu0(s)
E˚= E˚oxid + E˚red
= -(-0.762V) + (0.339V)
= 1.10V
(+) value for E˚ means rx is spontaneous
(-) value for E˚ means rx is NOT spont.
Cu0(s) + 2Ag1+(aq)
Cu2+(aq) + 2Ag0(s)
Relate E˚ and G˚
G˚ = -nFE˚
E˚ = standard cell potential
n = # e-s transferred
From balanced redox equation
F = Faraday’s constant
F = 9.648 x 104 C/mol e-
Relate E˚ and K
E˚
= RT (lnK)
nF
R = 8.31 J/mol K
T = Kelvin temp
n = # mol e-s transferred
F = Faraday’s constant
K = equilibrium constsnt
[conc] or (pp) prod/ [conc] or (pp) reactant
each raised to it’s coefficient as an exponent
E˚ = RT (lnK)
nF
At 25˚ C equation becomes
E˚
= (0.0257 V) ln K
n
If E˚ is (+) ln K is positive and K > 1, therefore
the forward reaction is favored
If E˚ is (-) ln K is negative and K < 1, therefore
the reverse reaction is favored
Nernst Equation
Voltage will increase with increases in
concentration
Voltage will decrease with decreases in
concentration
E = E˚- RT (ln Q)
n
At 25˚C the equation becomes
E = E˚- 0.0257 V (ln Q)
n
Electrolytic Cells
Process of electrolysis
A nonspontaneous reaction is made to occur
by pumping electrical energy into the system
Electrons are pushed from the anode to the
cathode by a direct current
Electroplating
Reduction of metal ions to metal atoms occurs
at the cathode
Can calculate the mass of metal deposited
through the pathway:
Coul
mol e-s
F
mol metal
stoic factor
mm of metal
(F = 96,480 C/mol e-)
g metal
Amperes x seconds = coulombs
Electroplating