Faraday’s Law: the amount of a substance produced or consumed in an electrolysis

reaction is

directly proportional to the quantity of electricity that flows through the circuit.
Electrical measurements:
Electric current – flow of electrons through an external current, I
- measured in units of Ampere (A)
Electric charge – quantity of electricity, Q
- the product of the current flowing through a circuit by the amount of time it flows
- measured in units of Coulomb (C)
the quantity of electricity that flows through a circuit in one second
Coulomb: if
the current is one ampere ie. 1 A = 1C/s
This relationship can also be expressed mathematically by:
Charge (coulomb) = current (in ampere) x time (in seconds)
or
Q = It

Exercises for Faraday’s Law

1. Calculate the minimum time, in seconds, required to deposit 40.0 g of copper at the
cathode of
an electrolysis cell containing CuSO4(aq) using a current of 20 000 mA.
2. How many minutes does it take to plate 0.925 g of silver onto the cathode of an
electrolytic cell
using a current of 1.55 A?
3. The nickel anode in an electrolytic cell decreases in mass by 1.20 g in 35.5 min. The
oxidation
half-reaction converts nickel atoms to nickel (III) ions. What is the constant current?
4. The following two half-reactions take place in an electrolytic cell with an iron anode
and a
chromium cathode.
Fe
(s) 6 Fe2+(aq) + 2eCr 3+

+ 3e- 6 Cr(s)
(aq)

During the process, the mass of the iron anode decreases by 1.75 g.
Find the change in mass of the chromium cathode.
5. A student wishes to set up an electrolytic cell to plate copper onto a belt buckle.
Predict the
length of time it will take to plate out 2.5 g of copper from a copper (II) nitrate solution
using a
2.5 A current. At which electrode should the belt buckle be attached?
6. Determine the mass of chlorine produced when a 200 A current flows for 24.0 h
through a cell
containing molten sodium chloride. At which electrode is the chlorine produced?
7. A trophy company is setting up a nickel plating cell using an electrolyte containing
nickel (III)
ions. Predict the current required to produce metal at a rate of 5.00 g/min.
8. In the electrolysis of a molten group II metal chloride 2.50 A of constant current is
passed
through a cell for 1.28 hours. Use the information provided below to determine the
identity of
the group II metal.
Mass of cathode before application of 25.720
 g
current: Mass of cathode after application of
30.949
current:
g

GALVANIC CELL /VOLTAIC CELL

Voltaic Cells
A Voltaic Cell (also known as a Galvanic Cell) is an electrochemical cell that uses
spontaneous redox reactions to generate electricity. It consists of two separate half-cells. A
half-cell is composed of an electrode (a strip of metal, M) within a solution containing
Mn+ions in which M is any arbitrary metal. The two half cells are linked together by a wire
running from one electrode to the other. A salt bridge also connects to the half cells. The
functions of these parts are discussed below.
Half Cells
Half of the redox reaction occurs at each half cell. Therefore, we can say that in each half-cell
a half-reaction is taking place. When the two halves are linked together with a wire and a salt
bridge, an electrochemical cell is created.
Electrodes
An electrode is strip of metal on which the reaction takes place. In a voltaic cell, the
oxidation and reduction of metals occurs at the electrodes. There are two electrodes in a
voltaic cell, one in each half-cell. The cathode is where reduction takes place and oxidation
takes place at the anode.
Through electrochemistry, these reactions are reacting upon metal surfaces, or electrodes. An
oxidation-reduction equilibrium is established between the metal and the substances in
solution. When electrodes are immersed in a solution containing ions of the same metal, it is
called a half-cell. Electrolytes are ions in solution, usually fluid, that conducts electricity
through ionic conduction. Two possible interactions can occur between the metal atoms on
the electrode and the ion solutions.
1. Metal ion Mn+ from the solution may collide with the electrode, gaining "n" electrons
from it, and convert to metal atoms. This means that the ions are reduced.
2. Metal atom on the surface may lose "n" electrons to the electrode and enter the solution as
the ion Mn+ meaning that the metal atoms are oxidized.
When an electrode is oxidized in a solution, it is called an anode and when an electrode is
reduced in solution. it is called a cathode.
Anode
The anode is where the oxidation reaction takes place. In other words, this is where the metal
loses electrons. In the reaction above, the anode is the Cu(s) since it increases in oxidation
state from 0 to +2.
Cathode
The cathode is where the reduction reaction takes place. This is where the metal electrode
gains electrons. Referring back to the equation above, the cathode is the Ag(s) as it decreases
in oxidation state from +1 to 0.
Remembering Oxidation and Reduction
When it comes to redox reactions, it is important to understand what it means for a metal to
be “oxidized” or “reduced”. An easy way to do this is to remember the phrase “OIL RIG”.
OIL = Oxidization is Loss (of e-)
RIG = Reduction is Gain (of e-)
In the case of the example above Ag+(aq) gains an electron meaning it is reduced. Cu(s) loses
two electrons thus it is oxidized.
Salt Bridge
The salt bridge is a vital component of any voltaic cell. It is a tube filled with an electrolyte
solution such as KNO3(s) or KCl(s). The purpose of the salt bridge is to keep the solutions
electrically neutral and allow the free flow of ions from one cell to another. Without the salt
bridge, positive and negative charges will build up around the electrodes causing the reaction
to stop.
Flow of Electrons
Electrons always flow from the anode to the cathode or from the oxidation half cell to the
reduction half cell. In terms of Eocell of the half reactions, the electrons will flow from the
more negative half reaction to the more positive half reaction.
Cell Diagram
A cell diagram is a representation of an electrochemical cell. The figure below illustrates a
cell diagram for the voltaic shown in Figure 1 above.
When drawing a cell diagram, we follow the following conventions. The anode is always
placed on the left side, and the cathode is placed on the right side. The salt bridge is
represented by double vertical lines (||). The difference in the phase of an element is
represented by a single vertical line (|), while changes in oxidation states are represented by
commas (,).
Constructing a Cell Diagram
When asked to construct a cell diagram follow these simple instructions.
Consider the following reaction:
2Ag+(aq) + Cu(s) ↔ Cu2+(aq)+ 2Ag(s)
Step 1: Write the two half-reactions.
Ag+(aq) + e- ↔ Ag(s)
Cu(s) ↔ Cu2+(aq) + 2eStep 2: Determine the cathode and anode.
Anode: Cu(s) ↔ Cu2+(aq)+ 2eCathode: Ag+(aq) + e- ↔ Ag(s)
Cu(s) is losing electrons thus being oxidized. Oxidation happens at the anode. Ag+ is gaining
electrons thus is being reduced. Reduction happens at the cathode.
Step 3: Construct the Diagram.
Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s)
The anode always goes on the left and cathode on the right. Separate changes in phase
by | and indicate the the salt bridge with ||.
Cell Voltage/Cell Potential
The readings from the voltmeter give the reaction's cell voltage or potential difference
between it's two two half-cells. Cell voltage is also known as cell potential or electromotive
force (emf) and it is shown as the symbol Ecell.
Standard Cell Potential: Eocell = Eoright(cathode) - Eoleft(anode)
The Eo values are tabulated with all solutes at 1 M and all gases at 1 atm. These values are
called standard reduction potentials. Each half-reaction has a different reduction potential,
the difference of two reduction potentials gives the voltage of the electrochemical cell. If
Eocell is positive the reaction is spontaneous and it is a voltaic cell. If the Eocell is negative,
the reaction is non-spontaneous and it is referred to as an electrolytic cell.