Electrochemistry is the chemistry of reactions that involve electron transfer. There are two known type of electrochemistry reactions such as spontaneous reactions and non-spontaneous reactions. In spontaneous reactions, electrons are released with energy that can be used in electrochemical cells. In non-spontaneous reactions, electrons have to be supplied with energy in order to produce chemicals wanted in electrolytic cells or electrolysis.
Usually there are always two electrodes present in which reduction and oxidation occurs. Oxidation occurs in within the anode while reduction occurs in the cathode.
Electrochemical Cells In the reaction
Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s)
Zinc powder reacts with copper nitrate in solution to form Zinc Nitrate and copper atoms, this reaction is known as a redox reaction. Energy is lost in the form of heat which makes this reaction a exothermic reaction.
half equations: Zn(s) → Zn2+ + 2e- Cu2+ + 2e- → Cu(s)
These half equations can occur in separate beakers if there is a travelling path for the electrons (via a wire) and ions (via a salt bridge).
eg. 2 beakers, the left one contains zinc nitrate (aq) and the right beaker contains copper nitrate (aq). The two beakers are connected with filter paper soaked in potassium nitrate which acts as a salt bridge for the ions to travel and also the two electrodes (zinc left copper right) are connected with a wire attached to a voltmeter.
Before the wire is connected, the two beakers are in equilibrium with their perspective metal ions and metal atoms. The position of equilibrium depends on the concentration and temperature of each solution.
Zn → Zn2+ + 2e- and Cu2+ + 2e- → Cu
When the wire is connected, electrons can successfully flow from the zinc to copper, leaving behind Zn2+ ions and forming CU atoms in the beaker. Potassium ions enter the CU beaker and nitrate ions enter the Zn beaker to keep the system electrically neutral. Each beaker is called a half cell and a series of cells connected is called a battery.
Standard electrode potentials Voltage produced by half cells depend on cells willingness to lose electrons and the other cell to gain electrons.
Cells connected with a voltmeter is measured in electromotive force (emf/Eo) and is used to determine oxidising and reducing agents.
Equilibrium reaction in each half cell is affected by concentration and power of their reagents.
Strengths of oxidising and reducing conditions such as concentration, temperature and pressure must be the same.
The standard half-cell A standard half cell contains H+/H2 to reduce the one half of reagents and oxidise the other half. EMF = 0
Standard conditions for emf measurements are: -Elements must be in pure form -Temperature must be +- 25o C -All concentrations should equal 1.0 mol L-1 -Pressure for gases 1.0 atm or 101.3 kPa
Platinum or graphite are often used as electrodes in electrochemical reactions.
Cell Diagrams Voltmeters give a positive or negative value, depending which way round the terminals of the voltmeter are connected. (you have to state which way round the readings were taken)
IUPAC is used to write cell diagrams.
eg. In the zinc/copper cell opposite, the zinc half-cell is on the left and the copper half cell is on the right.
Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)
- ‘//’ represents a salt bridge or device causing separation. - ‘/’ represents a change of phase, with the 2 phases in direct contact. - If the 2 species are in the same solution, they are separated with a comma. - The left hand electrode is written first - Left hand cell = oxidation (write reduced form first then the oxidised form) - Right hand cell = reduction (write oxidised form first then reduced form)
Electromotive force of a cell Standard electrode potentials (Eo) can be used to calculate the emf for a given cell
Eo(cell) = Eo(RHE) - Eo(LHE)
Swapped around electrodes will cause the sign of the voltage will be reversed. (positive becomes negative)
If Eo(cell) is positive, then the left side is experiencing oxidation and the right side is experiencing reduction. If Eo(cell) is negative, then the left side is experiencing reduction and the right side is experiencing oxidation.
Spontaneous reactions have a + Eo(cell).
eg. Calculate Eo(cell) and then write a overall cell equation.
Zn / Zn2+ // Cu2+ / Cu
Eo(Zn2+ // Zn) = -0.76 V Eo(Cu2+ / Cu) = +0.34 V
Eo(cell) = Eo(RHE) - Eo(LHE) = +0.34 V - (-0.76V) = +1.10 V
This answer of +1.10V tells us that oxidation is occurring on the left and reduction is occurring on the right.
Zn → Zn2+ + 2e- Cu2+ + 2e- → Cu
Zn + Cu2+--> Zn2+ + Cu
Predicting Reactions -Eo values are often used to predict whether a reaction is spontaneous or not. Several methods to solve these problems are:
A. Write cell diagram then use
Eo(cell) = Eo(RHE) - Eo(LHE)
B. Write half equations with their proposed Eo values (reverse sign of Eo for oxidation equation) then add 2 equations together to form one full equation.
Eo(cell) = Eo(Red) - Eo(Ox)
D.Follow patterns: half cell with the most negative Eo will be oxidised (oxidation number increases) while the half cell with the most positive Eo will be reduced (oxidation number decreases)
A is useful for cell diagrams & beakers + salt bridges B + C is used to decide whether a reaction will occur Method D is useful with a list of many half cells