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PbS (s) + 4H2O2 (aq) -> PbSO4 (s) + 4H2O (l)
PbS has been oxidized to PbSO4.
The oxidation state of sulfur increased from -2 in PbS to +6 in PbSO4.
H2O2 has been reduced to H2O.
The oxidation state of oxygen decreased from -1 in H2O2 to -2 in H2O.
Since oxidation and reduction occurred simultaneously in the reaction, it is a redox reaction.
(The reactant substance) has been oxidized / reduced to (the product substance).
The oxidation state of (the element) increased / decreased from (o.s. no.) in (the reactant substance) to (o.s. no.) in (the product substance).
Examples:
Explain whether the underlined substance has been oxidized or reduced, in terms of oxidation state.
1. ZnO (s) + C (s) -> Zn (s) + CO (g)
Zinc oxide has been reduced to zinc.
The oxidatation state of zinc decreased from +2 in zinc oxide to 0 in zinc.
2. PbS (s) + 4H2O2 (aq) -> PbSO4 (s) + 4H2O (l)
Lead(II) sulfide has been oxidized to lead(II) sulfate.
The oxidation state of sulfur increased from -2 in lead(II) sulfide to +6 in lead(II) sulfate.
3. PbO (s) + H2 (g) -> Pb (s) + H2O (g)
Lead(II) oxide has been reduced to lead.
The oxidation state of lead decreased from +2 in lead(II) oxide to 0 in lead.
4. 2FeCl2 (aq) + Cl2 (g) -> 2FeCl3 (aq)
FeCl2 has been oxidized to FeCl3.
The oxidation state of iron increased from +2 in iron(II) chloride to +3 in iron(III) chloride.
When ice melts, the heat energy gained from surroundings is used to weaken the bonds between particles. The movement of the particles changes from vibrating in fixed positions in the solid state to moving and sliding over one another freely in confined spaces in the liquid state. The arrangement of the particles changes from regularly arranged, tightly packed together in the solid state to irregularly arranged, close together in the liquid state.
The average kinetic energy of the particles remains constant, so temperature remains constant during melting.
A buffer is an aqueous solution that can resist significant changes in pH levels upon the addition of a small amount of acid or alkali.
There are two types of buffer solutions: acidic buffer and alkaline buffer.
Acidic buffer
Acid buffer solutions have a pH less than 7. It is generally made from a weak acid and one of its salts. Commonly used acidic buffer solutions are a mixture of ethanoic acid (a weak acid) and sodium ethanoate in solution (a weak basic salt), which have a pH of 4.76 when mixed in equal molar concentrations.
Alkaline buffer
Alkaline buffer solutions have a pH greater than 7 and are made from a weak base and one of its salts. A very commonly used example of an alkaline buffer solution is a mixture of ammonia (weak alkali) and ammonium chloride solution (a weak acidic salt). If these were mixed in equal molar proportions, the solution would have a pH of 9.25.
How do buffers work?
Buffers work by neutralizing any added acid (H+ ions) or base (OH- ions), by producing a weak acid or a weak alkali, respectively, hence maintaining the required pH.
For example, a buffer made up of the weak base ammonia, NH3 and its conjugate acid, NH4+. When HCl (strong acid) is added to this buffer system, the extra H+ ions added to the system are consumed by the NH3 to form NH4+, which is a weak acid. Now, because all the extra H+ ions are locked up and have formed a weaker acid, NH4+, thus the pH of the system does not change significantly. Similarly, when NaOH (strong base) is added to this buffer system, the NH4+ ion donates a proton to the base to become ammonia and water, which is a weak alkali, thus neutralizing the base without any significant pH change.
H2O (l) + NH3 (g) ⇌ OH– (aq) + NH4+ (aq)
Example:
Answer is A. The mixture of ammonia and ammonium chloride solution acts as an alkaline buffer solution, maintaining the pH at about 10.
An ionic equation is a balanced equation in which the substances are expressed as dissociated ions in aqueous solution. Always make sure the type of elements and the number of each type of elements are balanced on both sides of the equation. Equally important, but quite often left unchecked by students, is to make sure the net charge of the ions is balanced on both sides of the equation.
You can start off an ionic equation by first writing down the full chemical equation, or if it is a simple and common reaction, you can write down the ionic equation directly. (For simplicity, we assume an ionic equation means a net ionic equation, with the spectator ions omitted)
There are only 3 general rules to follow:
– If the substance is an ionic compound in the aqueous state, split it into its ions.
– If the substance is a strong alkali or a strong acid, split it into its ions.
– If it does not fall in categories 1 or 2, do not change anything. Leave it as it is.
Examples:
Precipitation Reaction:
AgNO3 (aq) + NaCl (aq) –> AgCl (s) + NaNO3 (aq)
Splitting into ions: Ag+ + NO3– + Na+ + Cl– –> AgCl + Na+ + NO3–
Cancel away the spectator ions: Ag+ + NO3– + Na+ + Cl– –> AgCl + Na+ + NO3–
Final ionic equation: Ag+ (aq) + Cl– (aq) –> AgCl (s)
Displacement Reaction:
Ca (s) + Cu(NO3)2 (aq) –> Ca(NO3)2 (aq) + Cu (s)
Splitting into ions: Ca + Cu2+ + 2NO3– –> Ca2+ + 2NO3– + Cu
Cancel away the spectator ions: Ca + Cu2+ + 2NO3– –> Ca2+ + 2NO3– + Cu
Final ionic equation: Ca (s) + Cu2+ (aq) –> Ca2+ (aq) + Cu (s)
Neutralisation Reaction:
2NaOH (aq) + H2SO4 (aq) –> Na2SO4 (aq) + 2H2O (l)
Splitting into ions: 2Na+ + 2OH– + 2H+ + SO42- –> 2Na+ + SO42- + 2H2O
Cancel away the spectator ions: 2Na+ + 2OH– + 2H+ + SO42- –> 2Na+ + SO42- + 2H2O
Final ionic equation, in simplest ratio: OH– (aq) + H+ (aq) –> H2O (l)
From Na to Si, the melting points increase, with a sharp increase from Na to Mg, and Al to Si, reaching a maximum for Si.
From Si to P, there is a sharp decrease in melting point, followed by a small increase in melting point from P to S, and a decrease in melting point from S to Ar.
Na, Mg, Al all have metallic structure. Moving from Na to Mg to Al, the number of protons and electrons increase, so there is an increasing electrostatic force of attraction between the positive ions and delocalised electrons, pulling the electrons closer to the nucleus, making the metallic bonding stronger. An increasing amount of energy is needed to break the increasing strength of the metallic bonds during melting. This results in the increasing trend of melting points from Na to Al.
Silicon exists as a giant molecular structure. A lot more energy is needed to break the large number of covalent bonds in the giant molecular structure, resulting in Silicon’s very high melting point.
Phosphorus, sulfur, chlorine and argon are non-metals which have simple covalent structures, with weak intermolecular forces of attraction. These forces require less energy to break during melting, resulting in their relatively lower melting points.
Phosphorus exists as P4 molecules, sulfur exists as S8 molecules, chlorine exists as Cl2 molecules and argon exists as individual atoms. The strength of the intermolecular forces of attraction decreases as the size of the molecule decreases, so melting points decrease from S to P to Cl to Ar. This explains the slight increase in melting point from phosphorus to sulfur, as sulfur molecules are bigger in mass than phosphorus molecules.
Diameter of an atom: 10-10 m
Diameter of Earth: 1.2742 x 107 m
Thickness of a strand of human hair: range from 17 x 10-6 m to 181 x 10-6 m
Length of a bus: 14 m
Diameter of a pin head: 1.5 x 10-3 m
Going down Group VII, the number of electronic shells increases, and atomic radius increases. As the valence electrons get further from the nucleus, the attractive force between the positive protons in the nucleus and the negative valence electrons decreases.
Hence, going down Group VII, it becomes more difficult for the atom to gain a valence electron to form an ion, causing reactivity to decrease.
Applying Fleming’s left-hand rule, the thumb, the index finger, and the middle finger on the left hand are positioned in such a way that they are pointing perpendicular to one another. The thumb will point in the direction of the force, the index finger will point in the direction of the magnetic field, while the middle finger will point in the direction of the conventional current flow.
This rule is also known as the Motor Rule. It is used in situations where electrical energy is converted to mechanical energy, similar to the energy conversion in a motor.
It is a process in which electrical charges are redistributed in a neutral body by bringing an electrically charged object near to it.
Note: This type of induction is most effective in electrical conductors, but can also happen in some electrical insulators.
Better still, test it out yourself! 🙂