Write an equation (chemical or ionic) and use oxidation numbers to find out what is being oxidized and what is being reduced.
(a) aluminum + oxygen
(b) magnesium + steam
(c) iron + dilute hydrochloric
(d) copper(ii) oxide + carbon
(e) bromine + potassium iodide solution
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To determine what is being oxidized and what is being reduced, we need to assign oxidation numbers and observe any changes in those numbers.
Oxidation numbers (also known as oxidation states) are assigned to individual atoms within a compound or molecule based on a set of rules. In general, the rules are as follows:
1. The oxidation number of an atom in a pure element is always zero.
2. The sum of the oxidation numbers in a neutral compound is zero.
3. The oxidation number of a monatomic ion is equal to its charge.
4. The most electronegative element in a compound is assigned its usual oxidation number.
5. Some common oxidation numbers: Hydrogen is +1, oxygen is usually -2, and fluorine is always -1.
Now, let's determine what is being oxidized and what is being reduced for each reaction:
(a) The reaction between aluminum and oxygen:
Aluminum is a pure element and has an oxidation number of zero.
Oxygen is diatomic, so its oxidation number is usually -2.
The equation for this reaction is:
2 Al + 3 O2 → 2 Al2O3
In this reaction, aluminum is being oxidized, as its oxidation number increases from zero to +3. Oxygen is being reduced, as its oxidation number decreases from -2 to -2.
(b) The reaction between magnesium and steam:
Magnesium is a pure element and has an oxidation number of zero.
Steam is a molecule consisting of two hydrogen atoms and one oxygen atom.
The equation for this reaction is:
Mg + H2O → Mg(OH)2 + H2
In this reaction, magnesium is being oxidized, as its oxidation number increases from zero to +2. Hydrogen in water is being reduced, as its oxidation number decreases from +1 to +1.
(c) The reaction between iron and dilute hydrochloric acid:
Iron is a pure element and has an oxidation number of zero.
Hydrochloric acid (HCl) dissociates into H+ and Cl- ions.
The equation for this reaction is:
Fe + 2 HCl → FeCl2 + H2
In this reaction, iron is being oxidized, as its oxidation number increases from zero to +2. Hydrogen in hydrochloric acid is being reduced, as its oxidation number decreases from +1 to 0.
(d) The reaction between copper(II) oxide and carbon:
Copper(II) oxide consists of copper with an oxidation number of +2 and oxygen with an oxidation number of -2.
Carbon is a pure element and has an oxidation number of zero.
The equation for this reaction is:
CuO + C → Cu + CO
In this reaction, copper in copper(II) oxide is being reduced, as its oxidation number decreases from +2 to 0. Carbon is being oxidized, as its oxidation number increases from zero to +2.
(e) The reaction between bromine and potassium iodide solution:
Bromine is a diatomic element and has an oxidation number of zero.
Potassium iodide dissociates into K+ and I- ions.
The equation for this reaction is:
Br2 + 2 KI → 2 KBr + I2
In this reaction, bromine is being reduced, as its oxidation number decreases from zero to zero. Iodine in potassium iodide is being oxidized, as its oxidation number increases from -1 to 0.
Remember, oxidation is the loss of electrons, while reduction is the gain of electrons.