redox and electrode potentials year 2 Flashcards
oxidising and reducing agents
- always be an oxidising agent and a reducing agent in redox reactions. Oxidising agent takes electrons from the species being oxidised. Oxidising agent contains the species that is being reduced
- reducing agent adds electrons to the species being reduced. The reducing agent contains the species that is oxidised.
eg 2Ag+ + Cu —> Cu2+ + 2Ag
Ag+ is the oxidising agent- Ag+ has oxidised Cu
Cu is the reducing agent- Cu has reduced Ag+
construction of redox equations using half equations
- only one element in half equations changes oxidation state
- half equation must balance the atoms
- half equations must balance for charge
Can construct equation by ensuring number of electrons on both sides balance and then joining the equations together, allowing the electrons to cancel. Can ensure electrons balance by multiplying each equation by a relative number to ensure electrons have the same coefficient in front
construction of redox equation using oxidation numbers
- use oxidation numbers to balance/ construct an overall equation of a redox reaction
- first work out the oxidation state of each of the atoms on both sides of the equation to find out which has been oxidised and which has been reduced
- then ensure oxidation changes balance. eg if the oxidation state of one element increases by +5 the oxidation state of the other element will need to decrease by -5. This can be achieved by placing large coefficient at the front of the molecules, therefore balancing the equation.
- Then finish balancing remaining molecules eg any H20
predicting products of redox reactions
- in equations you might not know all the species involved in the reaction and you might need to predict any missing reactants or products. In aqueous redox reactions, H20 is often formed. Other likely products are H+ and OH- ions depending on the conditions used. As a final check, ensure both sides of the equation are balanced by charge
eg when writing half equation for formation of Mn2+ from MnO4-, you need to add 5 electrons to the left hand side, to balance for charge, and then add water and hydrogen ions to further balance the atoms
redox titrations
- for redox titrations, the procedures are very similar to acid-base titrations
-common redox titrations: - potassium manganate (VII) (KMn04 (aq)) under acidic conditions
- sodium thiosulfate (Na2S2O3 (aq)) for determination of iodine (I2 (aq))
manganate (VII) titrations
- MnO4- ions are reduced and so the other chemical used must be a reducing agent that is oxidised
- the potassium manganate will be added to the burette.
- a measured volume of the solution being analysed will be added to the conical flask using a pipette. Excess of dilute sulfuric acid is also added to provide H+ ions required for the reduction of MnO4- ions. Reaction is self indicating
- manganate solution reacts and is decolourised as it is being added. End of titration occurs when the first permanent pink colour is seen, so Mn04- ions are in excess
- KMnO4 is a deep purple colour and is difficult to see the bottom of the meniscus through the intense colour, so burette readings are read from the top, rather than the bottom, of the meniscus
- can be used for the analysis of many reducing agents eg Fe+2 ions and ethanedioic acid (COOH)2
- manganate ions are often reduced to form Mn+2 ions
MnO4- + 8H+ + 5e- —–> Mn+2 + 4H2O - often the solution containing the reducing agent will be prepared in a volumetric flask, so will need to be scaled up by factors of 10 etc
- can also use for determination of a formula by working out the moles and Mr of an unknown compound, and then finding the formula eg the water of crystallisation number
- potassium manganate can be replaced with other oxidising agents eg acidified dichromate (VI). In this procedures and calculations will be similar
iodine/ thiosulfate redox titrations
- thiosulfate ions, S2O3 -2 are oxidised and iodine I2 is reduced (usually sodium thiosulfate that is used)
oxidation: 2 S2O3 -2 —–> S4O6 -2 + 2e-
reduction: I2 + 2e- —–> 2I-
Overall: 2S203 -2 + I2 —> 2I- + S4O6 -2 - the concentration of aqueous iodine can be determined by titration with a standard solution of sodium thiosulfate.
- add standard solution of sodium thiosulfate to burette
- prepare a solution of oxidising agent to be analysed and add this to a conical flask using a pipette. Then add an excess of KI. Oxidising agent reacts with iodide ions to produce iodine, which turns the solution a yellow-brown colour
- during the titration, the iodine is reduced back to I- ions and the brown colour fades gradually, making it difficult to decide on an end point. Use starch indicator when solution a pale straw colour. A deep blue-black colour forms. When this fades, after addition of more sodium thiosulfate, the blue-black colour will disappear as the iodine will have JUST reacted
examples of iodine-thiosulfate titrations
Different oxidising agents:
- chlorate (I) ions, ClO-
- copper (II) ions, Cu2+
- same principles applied, provided that the oxidising agents are capable of oxidising I- ions to I2
- ClO- is an active ingredient in household bleach.
ClO- + 2I- + 2H+ —> Cl- + I2 + H2O
2s2O3-2 + I2 —-> 2I- + S4O62- - Add bleach to a volumetric flask and make up to 250cm3. Then measure 25cm3 into a conical flask and add KI and HCl to acidify the solution. Then add sodium thiosulfate from the burette. Can determine the concentration of chlorate ions in the bleach, by working out the moles of S204 -2 that reacted and then determining the amount of I2 and ClO- , using the two separate equations. Then scale up and divide by the volume to find the concentration of ClO-
analysis of copper
iodine-thiosulfate titrations can be used to determine the copper content of copper(II) salts or alloys. Cu2+ ions are produced simply by dissolving the compound in water, for copper(II) salts
Insoluble copper(II) compounds can be reacted with acids to form Cu+2 ions
- for copper alloys, such as brass or bronze, the alloy is reacted and dissolved in concentrated nitric acid, followed by neutralisation to form Cu2+ ions
Cu(s) —> Cu2+ (aq)
- Cu+2 ions react with I- to form a solution of iodine and a white precipitate of copper iodide, CuI
2Cu+2 + 4I- –> 2CuI + I2
- the iodine is then titrated with a standard solution of sodium thiosulfate
- 2 mol Cu+2 produces 1 mol iodine which reacts with 2 mol S2O3 so 1 mol Cu2+ is equivalent to 1 mol S2O3 2-
electrochemical cells
- voltaic cell, type of electrochemical cell, converts chemical energy into electrical energy
- as electrical energy results from movement of electrons, need chemical reactions that transfer electrons from one species to another, redox reactions
half cells
- a half-cell contains the chemical species present in a redox half-equation. A voltaic cell can be made by connecting together two different half-cells, which then allows electrons to flow. In the cell, the chemicals in the two half-cells must be kept apart- if allowed to mix, electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy
metal/metal ion half-cells
- metal rod dipped into a solution of its aqueous metal ion. Represented using a vertical line for the boundary between the aqueous solution and the metal
- at the phase boundary where the metal is in contact with its ions, an equilibrium will be set up. The equilibrium in a half-cell is written so that the forward reaction shows reduction and the reverse reaction shows oxidation
Zn+2 +2e- <—-> Zn - in an isolated half-cell there is no net transfer of electrons either into or out of the metal
- when two half-cells are connected, the direction of electron flow depends upon the relative tendency of each electrode to release electrons
ion/ion half-cells
- contains ions of the same element in different oxidation states. eg a half-cell can be made containing a mixture of aqueous iron(II) and iron(III) ions
Fe+3 + e- <——> Fe+2 - no metal to transport electrons either into or out of the half-cell so inert metal electrode made out of platinum is used
electrode potentials
In a cell with two metal/ metal ion half cells connected, the more reactive metal releases electrons more readily and is oxidised:
In an operating cell:
- the electrode with more reactive metal looses electrons and is oxidised- this is the negative electrode
- the electrode with the less reactive metal gains electrons and is reduced- this is the positive electrode
standard electrode potential
- the tendency to be reduced and gain electrons is measured as a standard electrode potential.
The standard chosen is a half-cell containing hydrogen gas, H2, and a solution containing H+ ions. An inert platinum electrode is used to allow electrons into and out of the half-cell
Standard conditions are used, with a concentration of solutions of exactly 1mol/dm3 - the standard electrode potential is the e.m.f/ voltage of a half-cell connected to a standard hydrogen half-cell under standard conditions
- standard electrode potential of a standard hydrogen electrode is exactly 0V. The sign of a standard electrode potential shows the sign of the half-cell connected to the standard hydrogen electrode and shows the relative tendency to gain electrons compared with the hydrogen half-cell
measuring a standard electrode potential
- to measure a standard electrode potential, the half cell is connected to a standard hydrogen electrode
- two electrodes are connected by a wire to allow a controlled flow of electrons
- the two solutions are connected with a salt bridge which allows ions to flow. The salt bridge typically contains a concentrated solution of an electrolyte that doesnt react with either solution eg filter paper soaked in KNO3
- Standard electrode potentials have been measured for many different redox systems and are listed in data reference tables.
what happens the more negative the standard electrode potential value
- greater the tendency to LOSE electrons and undergo oxidation
- the LESS the tendency to gain electrons and undergo reduction
Metals tend to have negative standard electrode potential values and lose electrons
- the more negative the E value, the greater the REACTIVITY of a metal in loosing electrons
what happens the more positive the standard electrode potential value
- the greater the tendency to GAIN electrons and undergo reduction
- the LESS the tendency to lose electrons and undergo oxidation
Non metals tend to have positive standard electrode potential values and gain electrons
- the more positive the E value, the greater the reactivity of a non-metal in gaining electrons
cell potentials
- cells can easily be assembled using any half-cells. The e.m.f/ voltage measure is then a cell potential, Ecell.
- standard electrode potentials essentially quantify the tendency of redox systems to gain or lose electrons. A standard cell potential can be calculated directly from standard electrode potentials
Ecellꝋ = Eꝋ(positive electrode) - Eꝋ(negative electrode)
predicting redox reactions
- predictions can be made about the feasibility of any potential redox reactions using standard electrode potentials
- the most negative system has the greatest tendency to be oxidised and lose electrons
- the most positive system has the greatest tendency to be reduced and gain electrons
Oxidising AGENTS will be reduced and will be on the left.
A reducing agent adds electrons to the species being reduced so reducing agents are oxidised and are on the right
Can predict the feasibility of redox reactions using E values
a reaction should take place between an oxidising agent on the left and a reducing agent on the right, provided that the redox system of the oxidising agent has a more positive E value than the redox system of the reducing agent - the strongest reducing agent is at the top on the right
- the strongest oxidising agent is at the bottom on the left
When writing overall equations, half-equations must be combined. The oxidation half equation is obtained by reversing the equilibrium. The overall equation will have the reduction half equation plus the reversed oxidation equation
- redox system with more positive electrode potential will react from left to right and gain electrons
- redox system with the less positive electrode potential will react from right to left and lose electrons
limitations of using electrode potential values
- Reaction rate. Reactions have a very large activation energy, resulting in a very slow rate, so electrode potentials may indicate the thermodynamic feasibility but give no indication of the rate of a reaction
- Concentration. Standard electrode potentials are measured using concentrations of 1 mol/dm3. Many reactions take place using concentrated or dilute solutions. If the concentration of a solution is not 1mol/dm3 then the value of the electrode potential will be different from the standard value. Apply basic La Chatelliers principle rules to work out direction of shift based on increasing concentration
- Other factors. Actual conditions used may be different from the standard conditions used to record electrode potential values.
The standard electrode potentials apply to aqueous equilibria, but many reactions that take place arent aqueous
modern cells and batteries
- modern-day cells and batteries need to have two electrodes to have different electrode potentials
- divided into primary, secondary and fuel cells
primary cells
- non-rechargeable and are designed to be used once only. When in use, electrical energy is produced by oxidation and reduction at the electrodes. However, reactions cannot be reversed, so when chemicals used up the voltage will fall and the battery will go flat
- still used in low-current, long-storage devices eg clocks and smock detectors. Most are alkaline based on zinc and manganese dioxide. Zn/MnO2, and a potassium hydroxide alkaline electrolyte.
secondary cells
They are rechargable. The cell reaction can be reversed during recharging. The chemicals can be regenerated and the cell can be used again
eg lead-acid batteries in cars, lithium-ion batteries in laptops, tablets etc
IF asked to write overall equation when cell is being charged, just swap over the left and right hand side of the original equation
fuel cells
A fuel cell uses the energy from the reaction of a fuel with oxygen to create voltage
- fuel and oxygen flow into the fuel cell and the products flow out. The electrolyte remains in the cell
- Fuel cells can operate continuously provided that the fuel and oxygen are supplied into the cell
- fuel cells don’t have to be recharged
Many different fuels are used, but hydrogen is the most common. Hydrogen fuel cells produce no CO2 during combustion, and water is the only combustion product. Other fuels such as methanol are being developed for use in hydrogen fuel cells
hydrogen fuel cells
- a hydrogen cell can have either an alkali or acid electrolyte.
In an alkali hydrogen fuel cell: - hydrogen and oxygen both flow into the system, and are surrounded by an electrolyte of OH-
oxidation: H2 + 2OH- –> 2H20 + 2e-
reduction : 0.5O2 + H20 + 2e- —> 2OH-
overall: H2 + 0.5O2 —> H20
In an acid hydrogen fuel cell:
oxidation: H2 —> 2H+ + 2e-
reduction: 0.5O2 + 2H+ + 2e- —> H20
overall: H2 + 0.5O2 —> H20
