Chapter 23 - Redox and electrode potentials Flashcards
What species will always be in a redox reaction?
In a redox reaction, there will always be an oxidising agent and a reducing agent.
What is an oxidising agent and a reducing agent?
a. An oxidising agent takes electrons from the species being oxidised and the oxidising agent contains the species that is reduced.
b. A reducing agent adds electrons to the species being reduced and the reducing agent contains the species that is oxidised.
How would you write a half equation in alkaline conditions?
a. Firstly, work out the change in oxidation number which is equal to the number of electrons to be placed on the side of the equation with the greatest oxidation state.
b. After the electrons have been placed, balance the charge of the equation using OH- ions.
c. Then add H20 molecules to balance the oxygen atoms in the remaining equation.
How would you write a half equation in acidic conditions?
a. Balance the atoms on either side of the equation using H20 molecules and H+ ions only.
b. Then balance the overall charges on either side of the equation by using electrons.
How do Manganate (VII) titrations work?
- In Manganate (VII) [permanganate] titrations, MnO4- (aq) ions are reduced to Mn2+ ions and so the other chemical used must be a reducing agent that is oxidised.
a. A standard solution of known concentration of potassium manganate (VII) is added to the burette.
b. Using a pipette, add a measured volume of the solution being analysed to the conical flask. An excess of dilute sulfuric acid is also added to provide the H+(aq) ions required for the reduction of MnO4- (aq) ions. You do not need an indicator, as the reaction is self-indicating.
c. During the titration, the manganate (VII) solution reacts and is decolourised as it is being added. The end point of the titration is judged by the first permanent pink colour, indicating when there is an excess of MnO4- ions present.
d. Repeat the titration until you obtain concordant titres (two titres that agree within 0.10 cm^3)
How do you read the meniscus on Manganate (VII) titrations?
KMnO4 (aq) is a deep purple colour and is very difficult to see the bottom of the meniscus through the intense colour. So, burette reading are read from the top rather than the bottom.
How do you calculate the percentage purity of a compound?
Percentage purity = (mass worked out from the experiment [pure sample])/(mass of impure sample) x 100
How can the permanganate titration be extended to non-familiar redox titrations?
a. Permanganate titrations can be used to analyse other reducing agents that reduce MnO4- to Mn2+.
b. KMnO4 can be replaced with other oxidising agents, the commonest used being acidified dichromate (VI), H+ / Cr2072-.
How do iodine-thiosulfate titrations work and what can they be used to determine?
- Thiosulfate is oxidised from S2O32- to S4O62- and Iodine is reduced from I2 to I-.
- Iodine/thiosulfate titrations can be used to determine: The ClO- (oxidising agent) content in household beach, the Cu2+ (oxidising agent) content in copper (II) compounds and the Cu (oxidising agent) content in copper alloys.
a. Add a standard solution of Na2S2O3 to the burette.
b. Prepare a solution of the oxidising agent to be analysed. Using a pipette, add this solution to a conical flask. Then add an excess of potassium iodide. The oxidising agent reacts with iodide ions to produce iodine, which turns the solution a yellow-brown colour.
c. Titrate the solution with the Na2S2O3 (aq). During the titration, the iodine is reduced back to I- ions and the brown colour fades quite gradually, when the iodine colour has faded enough to become a pale straw colour, add a small amount of starch indicator. A deep blue-black colour forms to assist with the identification of the end point. As more sodium thiosulfate is added, the blue-black colour fades. At the end point, all the iodine will have just reacted, and the blue-black colour disappears.
What does it mean when the sulfur in the tetrathionate ion has a decimal oxidation state?
It just means that the sulfurs within the tetrathionate ion have different oxidation states.
In the iodine/thiosulfate reaction why is it important to only add the starch indicator towards the end of an iodine-thiosulfate titration?
a. The blue-black colour is a complex of starch and iodine. If added too early, the starch-iodine complex may precipitate out of solution, preventing some of the iodine reacting with the thiosulfate.
b. Close to the end point, the iodine concentration is low enough for the complex not to precipitate out.
Can the iodine-thiosulfate titration be extended to other non-familiar redox reactions?
The same principles can be applied for the analysis of other oxidising agents, provided that they are capable of oxidising I- ions to I2.
What are the ClO- ions commonly known as?
They are commonly known as hypochlorite.
How is the Cu2+ content in copper (II) compounds determined?
a. For copper (II) salts, Cu2+ (aq) ions are produced simply by dissolving the compound in water.
b. Insoluble copper (II) compounds can be reacted with acids to form Cu2+(aq) ions.
c. Cu2+ ions react with I-(aq) to form a solution of Iodine, I2(aq) and a white precipitate of Copper(I) iodide, CuI(s). The mixture appears as a brown colour.
d. The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate.
How is the Cu content in copper alloys determined?
a. For copper alloys, such as brass or bronze, the alloy is reacted and dissolved in concentrated nitric acid, followed by neutralisation to form Cu2+(aq) ions.
b. Cu2+ ions react with I-(aq) to form a solution of Iodine, I2(aq) and a white precipitate of Copper(I) iodide, CuI(s). The mixture appears as a brown colour.
c. The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate.
How do you calculate the percentage composition in an alloy?
% composition = mass of element/mass of alloy x 100
What is a voltaic cell?
a. A voltaic cell is a type of electrochemical cell which converts chemical energy into electrical energy.
b. As electrical energy results from movement of electrons, you need chemical reactions that transfer electrons from one species to another known as redox reactions.
c. A voltaic cell can be made by connecting together two different half cells which allow electrons to flow. A half-cell contains the chemical species present in a redox half-equation. The two half cells must be kept apart as if allowed to mix, electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy.
What is a metal/metal ion half-cell?
a. A metal/metal ion half-cell consists of a metal rod dipped into a solution of its aqueous metal ion. This is represented using a vertical line for the phase boundary between the aqueous solution and the metal.
b. At the phase boundary where the metal is in contact with its ions, an equilibrium will be set up.
i. By convention, the equilibrium in a half cell is written so that the forward reaction shows reduction and the reverse reaction shows oxidation.
c. When two half cells are connected, the direction of electron flow depends upon the relative tendency of each electrode to release electrons.
What is an ion/ion half-cell?
a. An ion/ion half-cell contains ions of the same element in different oxidation states.
b. In this type of half-cell there is no metal to transport electrons either into or out of the half-cell, so an inert metal electrode made out of platinum is used.
What is known as the standard electrode potential?
a. The standard electrode potential is the e.m.f. of a half-cell connected to a standard hydrogen half-cell under standard conditions of 298K, solution concentrations of 1 mol dm^-3, and a pressure of 100kPa.
b. The tendency to be reduced and gain electrons is measured as a standard electrode potential.
What is a standard hydrogen electrode?
a. A half-cell containing hydrogen gas, H2(g), and a solution containing H+ ions. An inert platinum electrode is used to allow electrons into and out of the half-cell. This is known as a standard hydrogen electrode.
b. By definition, the standard electrode potential of a standard hydrogen electrode is 0 V.
c. 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.
What are the standard conditions used in a standard hydrogen half-cell?
a. Solutions have a concentration of exactly 1 mol dm^-3.
b. The temperature is 298K (25 degrees celsius).
c. The pressure is 100kPa (1 bar).
How do you measure a standard electrode potential?
- To measure a standard electrode potential, the half-cell is connected to a standard hydrogen electrode.
a. The two electrodes are connected by a wire to allow a controlled flow of electrons.
b. 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 does not react with either solution.
c. One such example of a salt bridge is a strip of filter paper soaked in aqueous potassium nitrate, KNO3 (aq)
Within a voltaic cell, what does the standard electrode potential tell us?
a. The more negative the E value:
i. The greater the tendency to lose electrons and undergo oxidation.
ii. The less the tendency to gain electrons and undergo reduction.
iii. In general, the more negative the E value, the greater the reactivity of a metal in losing electrons.
b. The more positive the E value:
i. The greater the tendency to gain electrons and undergo reduction.
ii. The less the tendency to lose electrons and undergo oxidation.
iii. In general, the more positive the E value, the greater the reactivity of a non-metal in gaining electrons