Topic 10 - Oxidation and reduction
10.1 Oxidation and reduction
10.1.1 : Oxidation is the loss of electrons, reduction is the gain of electrons...ie Oxidation half equation... Mg -> Mg2+ + 2e-. Reduction half equation... O + 2e- -> O2-.
10.1.2 : The oxidation number of an element is zero. The oxidation number of an ion is equal to the charge of the ion. In compounds, Hydrogen has an oxidation number of +1. In compounds, Oxygen (usually...except in peroxides) has an oxidation number of -2 ... that should be enough information to deduce the oxidation numbers of every element in any IB reaction. The oxidation numbers of ions will be related to the group of the element (ie group 1 elements will be +1...up to group sever elements which will be -1). The elements in the elements in the d block will have multiple oxidation states.
10.1.3 : If an element is oxidized, its oxidation number will go up. If an element is reduced, its oxidation number will go down. To find out, simply write down the oxidation numbers for each element as explained previously.
10.1.4 : An oxidizing agent is an element which causes oxidation (and is reduced as a result). A reducing agent is an element which causes reduction (and is oxidized as a result).
10.1.5 : Iron Oxide, for example, come in two forms...FeO and Fe2O3. Iron (II) oxide and Iron (III) oxide respectively. The number after the Iron is determined by the oxidation number of the iron in the compound. This convention (including the use of roman numerals) is followed for all compounds in which there is some ambiguity as to the oxidation states of elements (generally d-block elements).
10.2 Electrolysis of a molten salt
10.2.1 : a diagram of an electrolytic cell should include a positive and negative electrode, a current source and the molten salt being electrolysed.
10.2.2 : Current is carried as in a normal circuit for the circuitry, and through the reaction occurring below. At the cathode positively charges ions gain electrons...ie X+ + e- -> X (the cathode must therefore be connected to the -ve pole of the source to supply electrons. At the anode negatively charged ions lose electrons...ie Y2- -> Y + 2e-. the circuit is completed by the two oxidation and reduction half equations creating an effectively complete circuit.
10.2.3 : In the above example, X will be produced at the cathode, and Y at the anode. Since only two species are present in a molten salt, there is no possibility for any other chemicals to be produced (ie water can not be involved etc...)
10.3 Reactivity
10.3.1 : A more reactive metal will displace a less reactive one from a compound and a more reactive halogen will displace a weaker one from a compound. This can be generalized to say a stronger reducing agent will displace a weaker one from a compound, and a stronger oxidizing agent will displace a weaker one from a compound. Thus, if a metal displaces another, we know it must be more reactive and ditto for halogens (which are the given examples).
10.3.2 : Based on the information in the IB data book...Add the two half equations in question together (one will have to be reversed, invert the sign of the E-zero value also) If the total E-zero value is positive, the reaction is possible. If not, it isn't. (This can be done based on their position in the table, but it makes more sense to use E-zero values, even if SL doesn't need to know them).