Chapter 23 Flashcards
What is an oxidising agent?
Substance that oxidises another atom or ion by causing it to lose electrons.
An oxidising agent itself gets reduced – gains electrons
What is a reducing agent?
A substance that reduces another atom or ion by causing it to gain electrons.
A reducing agent itself gets oxidised – loses/donates electrons
What side are electrons on a reduction half equation?
The left.
What side are electrons on an oxidation half equation?
The right.
How do you write a redox equation from half equations?
Firstly balance the electrons,
(if one has 2e- and the other has 4e- you would times everything in the first half equation by 2.)
Then add and cancel electrons as well as any other species that are on both sides of the equation.
How do you balance an equation using charges?
Write the imbalance equation and identify items that are changing in oxidation state. Deduce the oxidation state changes and then balance them. For example if magnesium goes from +7 to +2 and Iron and goes from +2 to+3 we would put a five in front of any species that has iron and one in front of any species with magnesium. You then calculate the charge on both sides of the equations an ad hydrogens to balance the charge. For example if the left side had a total charge of +9 and the right side had a total charge of +17 you would add 8H+ ions on the left hand side.
Describe how to carry out a Manganate (VII) titration and the results you would get
A standard solution of potassium manganate (KMnO4) is added to the burette, a measured volume of the solution being analysed (Fe) is added to the conical flask we also add an excess of dilute sulphuric acid to provide the H+ ions required for the reduction of MnO4- ions.
Manganate solution reacts and is decolorised as it’s being added the endpoint of the titration is judged by the first permanent pink colour bracket (this shows when there is an excess of MnO4 minus ions).
In this redox titration the manganate is the oxidising agent (Mg2+) and the iron is the reducing agent (Fe3+)
Why do we use dilute sulfuric acid in the manganate titration?
The acid must not react with the manganate ions, Sulfuric acid does not react with magnesium ions and doesn’t oxidise under these conditions.
Describe how to carry out a Iodine-thiosulfate titration and the results you would get
2S2O32– (aq) + I2 (aq) → 2I–(aq) + S4O62– (aq)
Thiosulfate ions are oxidised (S2O32-) and iodine I2 is reduced.
A standard solution f Na2S2O3 is added to a burette. The potassium oxide is added to iodide. which produces iodine which changed it a yellow brown colour.
Titrate this solution with Na2S2O3. The iodine is reduced back to I- and the brown colour fades quite gradually making it difficult to decide an end point. So a starch indicator is used when it becomes a pale straw colour.
A deep blue black colour forms which fades when more sodium thiosulfate is added (all the iodine has reacted)
How is an iodine thiosulfate titration used to determine the copper content of copper salts/alloys?
For copper alloys they are reacted and dissolved in concentrated nitric acid and then neutralised to form Cu2+ ions.
Cu(s) —> CU2+ (aq)
Cu2+ reacts with I- to form a solution of iodine and a white precipitate of copper iodide. (brown colour)
2Cu2+ + 4I- —> 2CuI + I2
The iodine in the brown mixture is titrated with a standard solution of sodium thiosulfate.
2S2O32- + I2 —> 2I- + S4O62-
2 moles of Cu2+ make 1 mol iodine which reacts with 2 mol S2O32-
How is an iodine thiosulfate titration used to determine the copper content of copper salts/alloys?
For copper alloys they are reacted and dissolved in concentrated nitric acid and then neutralised to form Cu2+ ions.
Cu(s) —> CU2+ (aq)
Cu2+ reacts with I- to form a solution of iodine and a white precipitate of copper iodide. (brown colour)
2Cu2+ + 4I- —> 2CuI + I2
The iodine in the brown mixture is titrated with a standard solution of sodium thiosulfate.
2S2O32- + I2 —> 2I- + S4O62-
2 moles of Cu2+ make 1 mol iodine which reacts with 2 mol S2O32-
What are elected potentials relative to?
The electrode potentials are measured relative to a standard hydrogen electrode
The standard hydrogen electrode is given a value of 0.00 V, and all other electrode potentials are compared to this standard
What are electrode potentials referred to as?
standard electrode potential (Eθ)
What is the standard electrode potential (Eθ)?
The potential difference ( sometimes called voltage) produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
What is the standard electrode potential (Eθ)?
The potential difference ( sometimes called voltage) produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
Is the more negative system oxidised or reduced?
oxidised and loses electrons
Is the more positive system oxidised or reduced?
Reduced and gains electrons.
Limitations of predictions using standard electrode potentials?
- electrode potentials may indicate the thermodynamic feasibility of a reaction but they give no indication of the rate of reaction.
-Standard electrode potentials are measured using concentrations of 1moldm-3 but sometimes we use solutions with different concentrations so the standard electrode potential isn’t accurate.
-the actual conditions may also differ from the standard conditions that the value are from
-they also apply to aq equilibria, many reactions occur that are not aqueous.
Is the cathode positive or negative?
negative- cations which are positively charged go to the cathode
Is the anode positive or negative?
positive - anions which are negatively charged go to the anode
Where does reduction occur?
At the cathode
Where does oxidation occur
At the anode
What are the 3 types of cell?
Primary, Secondary and Fuel.
What are primary cells?
Primary cells are non-rechargeable and can only be used once. Electrical energy is produced by oxidation and reduction but these reactions cannot be reversed so the chemicals will be used up, voltage will fall and the battery will go flat.
