Redox Flashcards
Oxidation definitions
gain of oxygen
loss of electrons
loss of hydrogen
increase in oxidation state
Reduction definitions
loss of oxygen
gain of hydrogen
gain of electrons
decrease in oxidation state
Oxidation number
the charge that an atom would have if all the covalent bonds were broken so that the more electronegative element kept all the electrons in the bond
Rules for oxidation states (numbers)
- elements = 0
- simple ions = same as charge
- in compounds:
- Oxygen = -2 (except in peroxides where O = -1, OF2 = +2)
- Hydrogen = +1 (except in metal hydrides = -1)
- Chlorine = -1 (except when with O or F)
- in polyatomic ions the oxidation number of the atoms add up to the net charge on the ion
- oxidation state of a transition metal in a complex ion can be worked out using the charges on the ligands and the overall charge
Oxidising agent (oxidant)
will oxidise other molecules but itself is reduced
Reducing agent (reductant)
will reduce other molecules but itself is oxidised
Steps in balancing a redox equation
- separate the redox equation into 2 half equations (omitting spectator ions)
- balance all atoms (except O and H)
- balance O using H2O
- balance H using H+ (the acid used is sulfuric acid as the SO4 2- ion does not react unlike other acids)
- balance the charges by adding electrons to the more positive side
- multiply each equation by some factor that will cause the electrons to cancel
- cancel like terms
Titration - Iron with manganate (VII)
this redox titration uses KMnO4 in an acidic solution as the oxidising agent, which oxidises Fe2+ ions into Fe3+
during the reaction MnO4 - is reduced to Mn2+
the reaction is accompanied by a colour change from purple to colourless so reaction acts as its own indicator
Titration - iodine-thiosulfate
several different titrations use an oxidising agent to react with excess iodide ions to form iodine
the excess iodine is then titrated with sodium thiosulfate using starch as an indicator
the starch indicator added during the titration forms a deep blue colour by complexing I2
as the I2 is reduced to I- during the reaction, the blue colour disappears
Winkler method for calculating dissolved oxygen
dissolves oxygen content of water is an important indicator of its quality (as pollution increases dissolves O2 generally decreases)
the biological oxygen demand (BOD) is used as a means of measuring the degree of pollution (BOD = amount of oxygen used to decompose the organic matter in a sample of water)
a high BOD indicates a lower level of dissolved O2
calculated based on a series of redox titrations
1. the dissolved O2 in the water is ‘fixed’ by the addition of manganese (II) salt. reaction of this salt with O2 in basic solution causes oxidation of Mn(II) to higher oxidation states such as Mn(IV)
2. acidified iodide ions (I-) are added to the solution, and are oxidised by the Mn (IV)
3. the iodine is then titrated with sodium thiosulfate
so for every 1 mol of O2 in the water 4 mol of thiosulfate ions are used
Voltaic cells
convert energy from spontaneous, exothermic chemical processes to electrical energy
oxidation occurs at the anode (negative electrode) and reduction occurs at the cathode (positive electrode)
they have a salt bridge (which contains a strong electrolyte - one that conducts electricity) these maintain electrical neutrality and completes the circuit
electrons flow from anode to cathode
Electrolytic cells
convert electrical energy to chemical energy, by bringing about non-spontaneous processes
oxidation occurs at the anode (positive electrode) and reduction occurs at the cathode (negative electrode) therefore cations migrate to the cathode and anions migrate to the anode
both electrodes are submerged in the same electrolyte and connected to a DC power supply
as the electric current passes through the electrolyte, redox reactions occur at the electrodes removing the charges on the ions and forming electrically neutral products
Cell diagram convention - voltaic cells
has the following features:
- a single vertical line represents a phase boundary such as that between a solid electrode and an aqueous solution within a half-cell
- a double vertical line represents the salt bridge
- the aqueous solutions of each electrode are placed next to the salt bridge
- the anode is on the left and the cathode on the right so electrons flow from left to right
- spectator ions are emitted
Standard hydrogen electrode (SHE)
the electrode potential of any two half cells are measured against the SHE
all reactions are completes under standard conditions
and under these conditions the standard electrode potential of the hydrogen electrode = 0 V
the SHE consists of hydrogen gas bubbled over a platinum electrode in contact wth 1 moldm-3 acid
Standard electrode potential
of a half-cell is its electrode potential relative to the standard hydrogen electrode (SHE), measured under standard conditions (pressure 100kPa, temp. 298K, all solutions 1 moldm-3)
all are reduction potentials and are measure of how much the reduction half equation wants to occur
when a half-cell is above H in the activity series e flow from half-cell to SHE, and electrode is given a -ve value
if the half-cell contains a metal below H in the activity series e flow from the SHE to the half-cell and the electrode has a +ve value
Anticlockwise rule
used to predict whether a redox reaction is thermodynamically able to occur
if you list the Eo values with the most -ve at the top then starting at the bottom left of a pair, the reaction proceeds anticlockwise (i.e the lower half reaction goes from left to right and the upper half goes from left to right)
the reaction is likely to go to completion if the Eo difference is > + 0.4 V
Redox couple
always written Eo ( oxidising agent I reducing agent) and redox couples always refer to reduction
Interpreting Eo values
the more +ve a Eo value the more that reduction half-equation wants to take place (the stronger the oxidising agent - itself is reduced)
the more -ve a Eo value the stronger the reducing agent (itself reduced)
electrons will always flow from the more -ve half-cell to the more +ve half cell
Eo cell = Eo red - Eo oxd
if Eo cell is +ve then reaction occurs spont.
if -ve then reverse reaction occurs spontaneously
never change the sign when subbing numbers into the formula
Eo cell and G
G = -nFEo
n - number of moles of electrons transferred
F - Faraday constant, the charge carried by one mole of electrons
when Eo = +ve, G = -ve => rxn is spont
when Eo = -ve, G = +ve => rxn is non-spont
when Eo = 0, G = 0 => rxn is at equilibrium
difficulties of electrolysis of aqueous solutions
makes predicting the products harder as water itself can be oxidised or reduced
- at the cathode:
2H2O + 2e- -> H2 + 2OH -
- at the anode:
2H2O -> 2H+ + O2 + 4e-
observations:
- colourless gases evolved at both electrodes
- ratio by volume of gases 2 H2: 1 O2
- the pH at the anode will decrease as H+ is released and pH at cathode will increase as OH- is released
so when solute M+A- is in an (aq) > 1 redox reaction pos. (competing reactions)
- at anode: either A- or H2O can be oxidised
- at cathode: either M+ or H2O can be reduced
How is outcome of electrolysis of aqueous solutions determined?
following factors:
- relative Eo values of the ions
- used for red. reactions as the more +ve Eo value will be reduced
- relative concentrations of the ions in the electrolyte
- used for oxd. of chlorine solutions; when Cl - conc. is low H2O is oxidised
- when conc. of Cl- is high (>25%) Cl- is preferentially oxidised
- the materials from which the electrodes are made
- relevant for electrolysis of CuSO4
Electrolysis of CuSO4 using C or other inert electrodes
at the cathode:
Cu 2+ is reduced as has higher Eo value
at the anode:
H2O is oxidised
observations:
- pink brown colour of copper deposited on cathode
- colourless gas O2 evolved at the anode
- decrease in pH as H+ released
- loss of blue colour due to loss of Cu2+
Electrolysis of CuSO4 using copper electrodes
at the cathode:
Cu2+ is reduced
at the anode:
Cu is oxidised
observations:
- pink brown colour of copper deposited on cathode
- disintegration of Cu anode
- no change in pH
- no change in intensity of blue colour as conc. of Cu2+ = constant
Electroplating
application of electrolysis
process of using electrolysis to deposit a layer of a metal
must have the following features:
- an electrolyte containing the metal ions which are to be deposited
- the cathode made of the object to be plated (reduction)
- sometimes the anode is made of the same metal which is to be coated because it may be oxidised to replenish the supply of ions in the electrolyte
reduction of metal ions at the cathode leads to their deposition on its surface
this process can be controlled by altering current and time according to how thick a layer of metal is desired
Factors affecting amount of product in electrolysis
charge = current X time
the charge carried by one mole of electrons = F
so using current and time charge can be found
then using F = Q/96500 the number of moles of electrons can be found
then using mole equation the mass of product can be found
