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CH1101 Tutorial 4 2014
Basic Electrochemistry
Professor Mike Lyons
melyons@tcd.ie
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Annual Examination 2010
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Use the following standard reduction potentials to calculate the equilibrium
Constant for the formation of the Zn(NH3)42+ ion.
Zn(NH3)42+(aq) + 2e-  Zn(s) + 4NH3(aq) E0 = -1.04 V
Zn2+ (aq) + 2e-  Zn(s) E0 = - 0.76 V.
The reaction of interest is: Zn2+(aq) + 4NH3(aq)  Zn(NH3)42+(aq).
This can be obtained by reversing the first half reaction and adding it to
the second.
Zn(s) + 4NH3(aq) + Zn2+(aq) + 2e-  Zn(s) + Zn(NH3)42+(aq) + 2eEcell0 = 1.04 V – 0.76 V = + 0.28 V
Now we recall K = exp[-G0/RT] = exp[nFEcell0/RT], lnK = nFE0cell/RT
lnK = (2 x 96485 Cmol-1x0.28 V)/(8.314 Jmol-1K-1x298K) = 21.8
K = e21.8 = 2.9x109.
K is large so reaction is very product favoured.
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Faraday’s Laws of Electrolysis.
The quantity (moles) of product
formed by the passage of an electric
current is stoichiometrically
equivalent to the amount (moles) of
electrons supplied.
The amount of product formed
during an electrolysis process is
calculated from the stoichiometry
of the reaction, the magnitude of
the current flowing, and the time
during which the current flows.
•
•
Michael Faraday : 1791-1867.
We will focus a lot on metal plating or metal
electrodeposition reactions.
Coulombs C
Mn+ + ne-  M
Amperes A Seconds s
Charge = current flowing x time taken
Charge passed by 1
mole electrons = 1F
= 96, 500 C.
n mol e- = 1 mol M
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Interfacial electron transfer at electrode/solution interfaces:
oxidation and reduction processes.
Electron sink electrode
(Anode).
Electron source electrode
(Cathode).
P
ne-
Q
Oxidation or de-electronation.
P = reductant (electron donor)
Q = Product
A
ne-
B
Reduction or electronation.
A = oxidant (electron acceptor)
B = Product
• The greater the applied voltage,
the larger the resulting current
flow, and the greater the rate
of the chemical reaction.
• The rate at which charge is
moved across the M/S interface
= the rate at which chemistry
is accomplished at the M/S
interface.
# electrons transferred
Current (A)
Electrode area (cm2)
dq
• In electrolysis we use an applied voltage
i
dt
to perform chemistry at a M/S interface.
• The applied voltage drives the chemical
reaction which does not occur spontaneously. Charge
(C)
• The current flowing across the M/S
interface is a measure of the rate of
Time (s)
the chemical transformation at the interface.
 nFA
dN
 nFAf 
dt
Amount of
Faraday Material (mol)
Constant (Cmol-1)
Reaction flux (rate)
mol cm-2s-1
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Electrolysis of a molten chromium salt for 1.5 hr with a constant
Current of 5.0 A resulted in the deposition of 4.835 g of chromium
Metal. Determine the charge on the chromium ion in the salt.
Our strategy is as follows. From the mass of Cr deposited, the amount
in mole of Cr deposited can be evaluated. The current passed and
deposition time can be used to evaluate the total quantity of charge
passed during electrolysis. The valency z of the Crz+ ion can be
evaluated by relating amount of Cr metal formed to amount of electrons
used.
Amount (mol) Cr: nCr = mCr/Acr = 4.835 g/51.996 gmol-1 = 0.093 mol
Amount charge passed Q = Ixt = 5.0 Cs-1 x 1.5 hr x 60 minhr-1 x 60smin-1
Q = 27,000 C.
1 mol electrons equivalent to charge of 96, 485 C (1 Faraday F)
Hence, 27,000 C equivalent to 27,000 C/96485 Cmol-1 = 0.28 mol
Hence 0.28 mol electrons produces 0.093 mol Cr metal
Hence we conclude that 1 mol Cr required 0.28 mol/0.093 mol = 3 electrons
Crz+ + ze-  Cr, 1 mol Cr = z mol eHence z = 3
And we write: Cr3+ + 3e-  Cr
A summary diagram for the stoichiometry of electrolysis
MASS (g)
of substance oxidized or reduced
M(g/mol)
AMOUNT (MOL)
of substance oxidized or reduced
AMOUNT (MOL)
of electrons transferred
balanced half‐
reaction
Faraday constant (C/mol e‐)
CHARGE (C)
time(s)
CURRENT (A)
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