Chemistry 30
Unit VIII – Electrochemistry
Electrochemistry
- the branch of chemistry dealing with the conversion of chemical energy to electrical energy (and vice
versa)
- spontaneous redox reactions – converting chemical potential  electrical energy
- non-spontaneous redox reactions – converting electrical energy  chemical potential
Galvanic/Voltaic Cells
- uses chemical potential energy to generate electrical energy (current)
- the redox reaction MUST be spontaneous
- used to make batteries (one or more galvanic cells in series – linked together)
- to allow a current to flow and for that current to be
useful, the two half reactions must be separated – into
two half-cells
- each half cell is made up of all of the parts necessary
for the half-reaction, and must include an electrolyte to
allow the ions to flow
- the example is a zinc – copper cell
- the anode is the site of oxidation – a zinc electrode
- it is immersed in a solution of zinc sulfate (the
sulfate ions are spectators)
- as the zinc is oxidized, it will lose electrons that will
flow through the wire (the external circuit) to the site
of reduction (from the anode to the cathode)
- the cathode is the site of reduction – in this case, a
copper electrode
- the copper electrode is immersed in a solution of copper (II) sulfate, the sulfate ions are
spectators
- as the electrons are pulled from the zinc (as it oxidizes) and travel through the wire (the external
circuit) to the copper electrode, these electrons will reduce copper (II) ions in solution to copper
metal, which will deposit on the cathode (the copper electrode)
- the salt bridge, which is made of spectator ions (like KNO3(aq)) forms the internal circuit
- the internal circuit is necessary to complete the loop and balance the charges in the half-cells
- as the electrons are lost from the anode, this half-cell becomes positively charged
- as the electrons are gained at the cathode, this half-cell becomes negatively charged
- the movement of anions (negative ions – in this case Cl-(aq) from the salt bridge and SO42(aq) from the CuSO4(aq) solution) to the anode balance the positive charge
- the movement of cations (positive ions – in this case, K+(aq) from the salt bridge and SO42(aq) from the ZnSO4(aq) solution) to the cathode balance the negative charge
- as the cell operates, a number of observations may be made:
- a cell potential (measured in volts, V) may be recorded on a voltmeter attached through the
external circuit
- the zinc electrode (which is being oxidized) will become lighter (have less mass)
- [Zn2+(aq)] will increase
- the copper electrode will become heavier (have more mass)
- [Cu2+(aq)] will decrease (the solution, which is blue, will become lighter in colour)
Some notes on constructing cells:
- although by convention we will always write the anode on the left and the cathode on the right, to
make a test question more difficult, the exam might have the ½ cells reversed – DETERMINE THE
SPONTANEOUS REACTION – that will identify the anode (oxidation) and cathode (reduction)
- if the substances being oxidized or reduced don’t contain a solid that could be used as an electrode,
use an inert electrode - either Pt(s) (platinum) or C(s) (carbon) – neither of these will react
Unit VIII – Electrochemstry
page 1
-
if a gas is required as a reactant, bubble the gas to the anode (or cathode) through a tube (shown
below left)
-
Instead of separating the half-cells in separate beakers and connecting them with a salt bridge, a
porous cup may be used instead – the cup allows the ions to flow through the solutions, but still
separates the cells adequately (see above right)
Cell Shorthand
- to describe a cell without drawing it, the following may be used (by convention, the anode is written
on the left, however to make the question more difficult, they may be reversed – ALWAYS CHECK
FOR SPONTANEITY!)
anode | anode electrolytes || cathode electrolytes | cathode
Include gases
needed here
-
Include all substances that
are produced/needed that
are in the solution
if the substances used are not at standard conditions (see the data booklet on p.9 – standards are
25C and 1.0 mol/L solutions), then the conditions must be included in the shorthand.
Ex: For the zinc copper cell in the example, where all solutions begin at 0.10 mol/L, the shorthand is:
Zn(s) | ZnSO4(aq) (0.10 mol/L) || CuSO4(aq) (0.10 mol/L) | Cu(s)
Ex: For a cell of nitrogen monoxide gas (with OH-(aq)) and Fe3+(aq) ions:
The half-reactions are found from p.9
Anode:
Cathode:
NO(g) + 2 OH-(aq)  NO2-(aq) + H2O(l) + eFe3+(aq) + e-  Fe2+(aq)
From these, write the cell shorthand:
Pt(s), NO(g) | OH-(aq), NO2-(aq), H2O || Fe3+(aq), Fe2+(aq) | C(s)
-
a platinum or carbon electrode may be used for both the anode and the cathode since neither
half-reaction contains a solid for an electrode
include the NO(g) with the anode
write all dissolved substances (including water) needed and produced in the half-reaction in the
electrolyte section.
Unit VIII – Electrochemstry
page 2
Standard Electrode Potentials
- a certain amount of electrical energy is produced/required when a half-reaction takes place – also
known as the electromotive force (EMF)
- just like heats of formation, they cannot be determined unless a reaction takes place, so all electrode
potentials must be relative to a specific standard – in redox, this standard is the reduction reaction
2 H+(aq) + 2 e-  H2(g)
E = 0.00 V (at 25C)
-
all other potentials are measured relative to this value.
From the hydrogen ½ cell standard, a cell potential (for a complete cell, both oxidation and reduction)
can be determined by setting up a cell with hydrogen as either the anode or the cathode (depending
on the strength of the other substance) and another substance. By measuring the overall cell
potential, the potential of the other substance may be calculated:
Ex: An unknown metal is placed in its cation solution and connected to a standard hydrogen half-cell. If
the metal is oxidized and the overall cell potential is 0.76 V, determine the oxidation potential of the
unknown metal, and identify the metal.
Since it is oxidized, we can write an oxidation ½ reaction for the unknown metal (since we don’t know
its charge, we can just use “n”, and the oxidation potential is unknown, let it be “x”:
E (V)
M(s)  Mn+(aq) + nex
2 H+(aq) + 2 e-  H2(g)
0.00
0.76 V
Since the overall potential is 0.76 V, and the hydrogen cell is 0.00V, then the unknown potential must
be 0.76 V. The table on p.9 lists the REDUCTION potentials of each ½ cell, not the oxidation
potentials – just like Hess’ Law, when you reverse a reaction, the sign of the electrode potential is
reversed. Given this, the unknown metal must be zinc.
**
The difference between Hess’ Law and calculating cell potentials is when we balance the reaction by
multiplying by a coefficient. In Hess’ Law, heats of formation are multiplied by a coefficient, since the
quantity of heat consumed or released is determined by the # moles of the substance. The EMF for a
substance is NOT affected by the number of moles since it is the amount of pull – the relative pull for
electrons is NOT affected by the number of atoms pulling. Therefore, you DO NOT need to balance
the equation and multiply by the coefficients to determine the cell potential.
-
Cell potentials may be calculated easily using the values on p.9
Ex: What is the cell potential of the following cell: Ag(s) | Ag+(aq) || Cr2+(aq) | Cr(s)
First, determine the two reactions – remember it must be a spontaneous reaction (in this case, they
switched the convention and wrote the cathode reaction first)
Anode: Cr(s)  Cr2+(aq) + 2e
Cathode: Ag+(aq) + e  Ag(s)
Then look up the E values on p. 9, remembering to change the sign on the anode (oxidation)
reaction
E (V)
Anode: Cr(s)  Cr2+(aq) + 2e
0.91
Cathode: Ag+(aq) + e  Ag(s)
0.80
Overall:
-
1.71 V
The positive value for E for the cell means that the reaction is spontaneous and will generate
electrical energy
Unit VIII – Electrochemstry
page 3
Other types of problems involving cell potential
1. The reference cell is the hydrogen half cell, but sometimes they will ask you to reset it:
Ex: what is the reduction potential of an iodine-iodide ion half-cell if the reference cell were sodiumsodium ion instead of hydrogen?
Since the reference point is now Na(s)  Na+(aq) + e, the entire table has been shifted up by the
difference between this ½ cell and the hydrogen ½ cell – a difference of 2.71V. You must add 2.71 V
to the Na(s)  Na+(aq) + e to get it to 0.00V, so you would add that to every ½ reaction on the table.
I2(s) + 2 e  2 I-(aq) would now be 0.54 + 2.71 = 3.25 V
If they asked for the oxidation potential instead, you have to remember to reverse the reaction, and
therefore change the sign of the potential.
Ex: what is the oxidation potential of an cobalt-cobalt(II) ion half-cell if the reference cell were sodiumsodium ion instead of hydrogen?
The difference would still be 2.71 V, so add 2.71 V to the reaction they want, which is:
Co2+(aq) + 2e  Co(s), so E for this reaction is –0.28 + 2.71 = 2.43 V; but since they want an
oxidation potential instead, flip the reaction and change the sign, so the oxidation potential is –2.43 V.
2. Determining the potential of an unknown:
Ex: (from January 1997 diploma) A NiCad battery has an electrical potential of 1.20 V. The halfreactions are:
Anode:
Cathode:
Cd(s) + 2 OH-(aq)  Cd(OH)2(aq) + 2e
NiOOH(s) + H2O(l) + e  Ni(OH)2(aq) + OH-(aq)
E = -0.49 V
Determine the reduction potential for the first half-reaction.
Since we know the overall potential, set up the reactions as you know them:
E(V)
Anode:
Cd(s) + 2 OH-(aq)  Cd(OH)2(aq) + 2e
x
Cathode:
NiOOH(s) + H2O(l) + e  Ni(OH)2(aq) + OH-(aq)
-0.49
Overall:
1.20 V
*** Therefore, 1.20 = -0.49 + x, so x = 1.69 V.
*** BUT (and a BIIIIG but) they asked for the REDUCTION potential of the first reaction, so make sure to
flip the reaction and change the sign. The correct answer is –1.69V
Applications of Spontaneous Redox Reactions
Batteries
- electrochemical (voltaic, galvanic) cells that are used in power generation
- a combination of two chemicals will always produce the same electrical potential (voltage)
- to increase voltage, hook more cell together in series (two or more cells in series is a battery)
Unit VIII – Electrochemstry
page 4
Primary Cells (dry cells)
- non-rechargeable
- electrolytes are a paste instead of a liquid
- most common – Lelanché cell – manganese oxide &
zinc, also Zn-HgO2 (get the ½ reactions from p.9)
LeLanché Cell
E (V)
0.76
1.22
A: Zn(s)  Zn2+(aq) + 2e
C: MnO2(s) + 4H+(aq) + 2e  Mn2+(aq) + 2 H2O(l)
Net:Zn(s) + MnO2(s) + 4H+(aq) Zn2+(aq) Mn2+(aq) + 2 H2O(l) 1.98 V
Secondary Batteries (storage batteries)
-
-
rechargeable
can be very small to very large
includes NiCd, lead-acid (cars), Ni-MH, Li-ION
car battery = 6 2V cells connected in series
to recharge a secondary battery (or cell), run the
current through the battery in the opposite direction
(reverses both of the half-reactions)
BUT if the quantity of H2SO4(aq) gets too low, the
battery cannot be recharged
Lead Acid Battery:
A: Pb(s) + SO42-(aq)  PbSO4(s) + 2e
C: PbO2(s) + SO42-(aq) + 4 H+(aq) + 2e  PbSO4(s) + 2 H2O
E (V)
0.36
1.69
Overall: Pb(s) +2SO42-(aq) PbO2(s) + 4 H+(aq) 2 PbSO4(s)
2.05 V
Fuel Cells
- changes chemical energy from a combustion reaction directly into electrical energy - no turbines
necessary – as are presently used in power-generating stations to create electricity from burning coal
- a much more efficient way of creating electricity from “combustion” than through burning
- requires the continuous addition of reactants/products, involves no toxic waste
- usual fuels – oxygen & hydrogen, also the methanol (CH3OH(l)) fuel cell
Hydrogen Fuel Cell
A: H2(g) + 2 OH-(aq)  2 H2O(l) + 2e
C: O2(g) + 2 H2O(l) + 4e  4 OH-(aq)
E (V)
0.83
0.40
Overall: 2 H2(g) + O2(g)  2 H2O(l)
1.23 V
Unit VIII – Electrochemstry
page 5
Corrosion
- although this doesn’t seem like an “application” of spontaneous redox reactions, the corrosion of
metal behaves just like a voltaic cell: - the reaction between purified metals exposed to oxygen and
water
- will occur to any metal that is a better reducing agent than OH-(aq) as long as three necessary factors
are present:
- water
- oxygen
- electrolyte – there must be at least some electrolyte present, however the more electrolyte, the
faster rusting will occur.
The pure (exposed) metal acts as the anode – ex: iron
Water and oxygen together act as the cathode
A: Fe(s)  Fe2+(aq) + 2e
C: O2(g) + 2 H2O(l) + 4e  4 OH-(aq)
**
The Fe2+(aq) formed isn’t what we call “rust” – to form Fe2O3(s), what we call “rust”, two additional
steps must take place (you’re not responsible for these, but for interest sake…)
When Fe2+(aq) is exposed to oxygen,
4 Fe2+(aq) + 2 O2(g) + 2 H2O(l)  2 Fe3+(aq) + 4 OH-(aq)
2 Fe3+(aq) + 6 OH-(aq)  Fe2O3(s) + 3 H2O(l)
What we call rust.
-
To demonstrate the rusting process, nails may be submerged in an agar gel that contains two
indicators, the Fe(CN)63-(aq) ion and phenopthalein.
- Fe(CN)63-(aq) turns blue in the presence of Fe2+(aq), showing the oxidation of the metal
- Phenolpthalein turns pink in the presence of base (OH-(aq)) showing the cathode – reduction of
oxygen
The following drawings demonstrate the results after only a few days:
Fig.1
Fig.2
Fig.3
Fig.4
Figure 1 – illustrates the normal rusting process. The ends of nails which are more stressed than the
middle (due to pounding and pushing into boards) will tend to corrode (and therefore show signs of
oxidation – the blue colour). The middle of the nail is showing signs of reduction – the presence of
the base (OH-(aq))
Figure 2 – shows how rusting can occur when other parts of the nail are stressed – in this case, the nail
was bent first, and shows signs of rusting in the middle as well as on the ends.
*** Nails on a deck or on a fence will corrode under the board – since the head is exposed to the water
and oxygen, this part of the nail will act as the cathode, and be reduced – not deteriorating the iron.
Unit VIII – Electrochemstry
page 6
The nail underneath will act as the anode, donating electrons – your deck nails are corroding
underneath and you can’t even tell!!!
Figure 3 – shows the effect of connecting the iron nail to a metal such as copper, which is a weaker RA
than iron. The iron corrodes even more than without the copper. In this situation, the iron is behaving
as a sacrificial anode. The exposed copper, which would normally corrode itself, is being protected
by the iron since it is SRA – it is oxidized instead of the copper.
Figure 4 – shows the effect of connecting the iron nail to a metal (in this case zinc) which is SRA than
iron. In this case, the iron is protected by the zinc, which is acting as a sacrificial anode. The zinc is
donating electrons instead of the iron, protecting the iron from corrosion.
So, now that you know all of this, how could you protect a metal from corrosion?
Protection from Corrosion
- Protect the metal from exposure to water and oxygen
- Paint it
- Coat it with oil or wax
- Coat with plastic
- Use a metal that is too weak an SRA to react with water – jewellery is made from Ag(s), Au(s)
because they do not react spontaneously (but rather $$)
- Plate the cheaper metal with a weak RA like Au(s) or Ag(s). This is much cheaper, but seals the
metal that would normally corrode from water and oxygen
- Mix the metal into an alloy (by mixing in 5% chromium, this reduces rusting by 80%, at 16%
chromium, this almost eliminates rusting)
- Coat the metal with a reactive metal (?!?) like zinc – called galvanization
- When some metals (like zinc, aluminum, tin, copper, nickel) corrode, they form a protective
barrier against further corrosion
- Connect the metal (like Fe) to a stronger RA like Zn(s) or Mg(s). The SRA will act as a sacrificial
anode, corroding instead of the metal you want to protect
- Connect the metal (like Fe) to a battery – the battery will supply the electrons instead of the metal –
the battery could be recharged using solar panels – this is called “cathodic protection”
Electrolytic Cells
So far, we have been interested only in spontaneous redox reactions and their “uses”. In this section, we
will be covering the uses of non-spontaneous reactions.
Electrolysis
- When electrical energy is passed through a solution containing ions (or a molten ionic compound) the
energy may be used to power a non-spontaneous redox reaction  electrolytic cell
- electroylic cells have many different constructions depending on the application, but the general
electrolytic cell is shown on the left.
Similarities to galvanic cells:
- there is both an anode and a cathode
- the electrons flow from the anode to the cathode via the external circuit
- the cations move to the cathode, anions to the anode
- there must be an electrolyte, wire, beaker
- the likely reaction will still be the SOA and SRA (but there are
exceptions – you must know one – the chlor-alkali cell)
Differences:
- there is no need to separate the anode and the cathode ½ cells (but
sometimes they are…)
- a power source (a battery in this diagram) is required to force the
nonspontaneous reaction to happen (no current is generated)
- the net cell potential is negative!
Unit VIII – Electrochemstry
page 7
You must be able to:
- identify an electrolytic cell and label its parts
- draw an electrolytic cell and its parts, including the direction of flow of electrons and ions
- predict the likely reaction
- write the likely reaction including the predicted required voltage
- use observations to confirm a reaction
- perform calculations to determine quantities of substances produced/consumed during the operation
of a cell (galvanic cells too).
Ex: Predict the half-reactions and the likely overall reaction and required voltage of an electrolytic cell
containing aqueous sodium iodide.
Na+(aq)
OA
I-(aq)
SRA
H2O(l)
SOA/RA
Just like when predicting any other redox reactions, list all of the
species and determine the SOA and SRA. The major difference
between this and galvanic cells is that they won’t give you as many
materials, you won’t have metals and the metal ions for all species
present. Often the electrodes are made of platinum or carbon
A: 2I-(aq)  I2(s) + 2 e
C: 2 H2O(l) + 2e  H2(g) + 2OH-(aq)
E(V)
-0.54
-0.83
Net: 2 H2O(l) + 2I-(aq)  H2(g) + 2OH-(aq) + I2(s)
-1.37 V
Once you have the SRA and SOA, write the half-reactions – the
oxidation ½ reaction is always the anode, the reduction is the
cathode. When you look them up on p.9, it is clear that the
reaction is non-spontaneous (the OA is lower than the RA, and
when you add the ½ cell potentials, the overall is negative)
This reaction is non-spontaneous and would required an input of 1.37 V. Evidence of this reaction would
include an appearance of a brown solution (dissolved iodine is brown), an increase in pH (the production
of OH-(aq)) and bubbles (H2(g))
*** These reactions are theoretical only – some Na+(aq) may also reduce.
*** Water is often the SRA or SOA (sometimes even both!)
Ex: Predict the likely reactions in the electrolysis of aqueous sodium sulfate.
Na+(aq)
OA
SO42-(aq)
RA
H2O(l)
SOA/SRA
In this case, water is going to act as both the SOA and SRA – when
the same substance is both the OA and RA (both oxidized and
reduced) this is called DISPROPOTIONATION
A: 2 H2O(l)  O2(g) + 4H+(aq) + 4e
C: 2 H2O(l) + 2e  H2(g) + 2OH-(aq)
E(V)
-1.23
-0.83 X 2
Net: 6 H2O(l)  2 H2(g) + O2(g) + 4H+(aq) + 4 OH-(aq)
-2.06V
This reaction commonly takes place in a “Hoffman’s
Apparatus”.
- Although the electrolyte (which is the sodium sulfate)
spans both the anode and the cathode, the anode and
cathode collect separately.
- At the anode, oxygen gas is collected (which can be
tested with a glowing splint that will relight), and as
hydrogen ions are produced, blue litmus paper will turn
red
- At the cathode, twice as much hydrogen gas is collected
(which will pop with a lit splint), and since hydroxide ions
are produced, red litmus paper will turn blue.
Unit VIII – Electrochemstry
page 8
The Chlor-Alkali Cell/ Electrolysis of Brine (NaCl(aq))
(The Exception you MUST know)
In Fort Saskatchewan (outside Edmonton) there are huge salt caves. Water is pumped into the caves to
dissolve the salt, and then the salt water (brine) is separated using electrolysis.
Ex: Predict the likely reactions in the electrolysis of NaCl(aq) (brine)
Na+(aq)
OA
Cl-(aq)
RA
In predicting this reaction, water is both the SOA and SRA…. So the likely reaction is
water being oxidized and reduced, but this is not the case…
H2O(l)
SRA/SOA
p.9
E(V)
Cl2(g) + 2e  2 Cl-(aq)
1.36
O2(g) + 4H+(aq) + 4e  2 H2O(l)
1.23
SRA
The Cl-(aq) ½ reaction is fairly close in potential to the H2O(l) ½ reaction and in reality
the water ½ reaction is SOOOOOO slow that IF (and only if) there is enough electrical
energy provided, even though the SRA is water, the FRA (faster reducing agent) is Cl (aq) and this reaction will actually occur….. this is called OVERVOLTAGE (if enough
energy is provided, a weaker OA or RA will react if this reaction is faster)
So the reactions are, if enough energy is provided:
SOA
2 H2O(l) + 2e  H2(g) + 2 OH-(aq)
-0.83
Na+(aq) + e  Na(s)
-2.71
Very little gas
will be collected
since Cl2(g) is
highly soluble
Chlorine
bubbles
E(V)
-1.23
-1.36
-0.83
A: 2 H2O(l)  O2(g) + 4H+(aq) + 4e
A: 2 Cl-(aq)  Cl2(g) + 2e
C: 2 H2O(l) + 2e  H2(g) + 2 OH-(aq)
Net: 2 H2O(l) + 2 Cl-(aq)  H2(g) + 2 OH-(aq) + Cl2(g)
-2.19 V
This reaction can also be demonstrated in a “Hoffman’s
Apparatus”.
- This reaction will produce the same products at the
cathode as with sodium sulfate
- The anode reaction is confirmed by the presence of
chlorine gas as a product (when tested with litmus
paper, it bleaches the paper and you can smell the
chlorine gas)
Other Uses of Electrolytic Cells
- Electroplating
- Deposits a thin uniform layer of metal on a conducting surface – you
can plate non-conducting materials by treatin them with wax or a
graphite laquer
- Used to plate gold on copper jewellery, make silver plated cutlery,
create “bronze” baby shoes, etc.
- If you want to plate copper, use a solution of CuSO4(aq), place the
object you want to plate at the CATHODE (in this case a fork).
- From the contents, Cu2+(aq), SO42-(aq), H2O(l), the SRA will be H2O
and the SOA will be Cu2+, which will cause the copper(II) ion to be
reduced at the cathode (plated onto the cathode).
- Purifying metals – electrolytic purification
- By choosing the voltage carefully, you can sequentially plate metals out of a solution starting with
the metal requiring the least voltage to plate (the metals will always be plated on the cathode)
Unit VIII – Electrochemstry
page 9
-
The Down’s Cell (the separation of MOLTEN sodium chloride (NaCl(l))
- Since Na+ is such a weak OA, it will never plate from an aqueous solution, so to obtain pure
sodium, water must not be available as an OA. To plate Na(s), start with molten (melted) sodium
chloride so that the only substances present are Na+(l) and Cl-(aq), then Na+ will be reduced at
the cathode and Cl- will be oxidized at the anode.
Electron Stoichiometry (or Quantitative Electrolysis)
Since ½ reactions are balanced in mole:mole ratios with chemical substances and electrons,
stoichiometric calculations may be performed using moles of electrons (strange but true) and Faraday’s
laws.
Faraday’s laws
- The amount of product/reactant produced/consumed is proportional to the time a circuit is in
operation
- For 1 mol of a product, 9.65x104 C is required per electron per mol of product (this is the conversion
factor for mol of e and mol of charge)
A couple of definitions:
Charge (q) – the electrical charge (measured in Coulombs, “C”);
- the charge on 1 electron is 1.60x10-19C
- the charge per mol of electrons is 9.65x104 C (the conversion unit is 9.65x104 C/mol e)
Current (I) – the amount of electricity through the cell per second (measured in Amperes, A = C/s)
- the conversion unit you will use is the value for current in C/s (Coulombs per second)
To solve these problems, always use a balanced equation that includes electrons! (Half-reaction only!)
Ex: Calculate the mass of copper plated from a solution of CuSO 4(aq) if the cell is run for 75 min. at a
current of 1.75A.
Cu2+(aq) + 2e  Cu(s)
-
As in all stoichiometry, start with a balanced equation – whenever you see time and current, it
will always be a ½ reaction involving whatever substance they mention.
Some of the conversion units are new, but they still work – always change A to C/s so you can cancel
the time, and use 9.65x104C/mol e to get to moles of electrons
Write A as C/s to
cancel the time
Use the conversion
unit to get to moles of
electrons
Convert time to s
xgCu (s ) 
1.75C 75 min 60s
1mole 
1molCu 63 .55 gCu





s
1min 9.65  10 4 C 2mole 
1molCu
 2.6gCu (s )
Molar ratio of electrons to
Cu from the ½ reaction
Unit VIII – Electrochemstry
page 10
Ex: What current must be applied to plate 0.755 g of silver from a solution of AgNO3(aq) in 10.0 min?
Ag+(aq) + e  Ag(s)
xC 0.755 gAg
1molAg
1mole  9.65 x10 4 C
1
1min







s
107 .87 gAg 1molAg
10 .0 min 60 s
mole
 1.13
Current is in A or
C/s (use C/s to
cancel units)
C
s
 1.13 A
Unit VIII – Electrochemstry
You need time on
the bottom
page 11