Edexcel A Level Chemistry:复习笔记2.1.4 Ionic Equations

Writing Chemical Formulae

 

  • Oxidation numbers are a useful tool for naming compounds as some elements can exist with more than one oxidation number
  • For compound with two elements it is straight forward to name the compound
  • For example
    • PCl3 is phosphorus(III) chloride or phosphorus trichloride
    • PCl5 is phosphorus(V) chloride or phosphorus pentachloride
    • OF2 is oxygen difluoride
    • O2F2 is dioxygen difluoride
  • In order to name a more complete compound we use Roman numerals for the element that has a variable oxidation number
    • K2CrO4 potassium chromate(VI)

Worked Example

Can you name these metal compounds?

  1. Cu2O
  2. MnSO4
  3. Na2CrO4
  4. KMnO4
  5. Na2Cr2O7

Answer:

Answer 1: copper(I) oxide:

The ox. no. of 1 O atom is -2 and Cu2O has overall no charge so the ox. no. of Cu is +1

Answer 2: manganese(II) sulfate:

The charge on the sulfate ion is -2, so the charge on Mn and ox. no. is +2

Answer 3: sodium chromate(VI):

The ox. no. of 2 Na atoms is +2 so CrO4 has an overall -2 charge, so the ox. no. of Cr is +6

Answer 4: potassium manganate(VII):

The ox. no. of a K atom is +1 so MnO4 has overall -1 charge, so the ox. no. of Mn is +7

Answer 5: sodium dichromate(VI):

The ox. no. of 2 Na atoms is +2 so Cr2O7 has an overall -2 charge, so the ox. no. of Cr is +6. To distinguish it from CrO4 we use the prefix di in front of the anion

Ionic Half-Equations

Balancing full ionic equations

  • Balancing equations using redox principles is a useful skill and is best illustrated by following an example
  • It is important to follow a methodical step-by-step approach so that you don't get lost:

Worked Example

Writing overall redox reactions

Manganate(VII) ions (MnO4- ) react with Fe2+ ions in the presence of acid (H+) to form Mn2+ ions, Fe3+ ions and water

Write the overall redox equation for this reaction

Answer

Step 1: Write the unbalanced equation and identify the atoms which change in oxidation number

1.6-Electrochemistry-Step-1-Writing-overall-redox-reactions

Step 2: Deduce the oxidation number changes

1.6-Electrochemistry-Step-2-Writing-overall-redox-reactions

Step 3: Balance the oxidation number changes

1.6-Electrochemistry-Step-3-Writing-overall-redox-reactions

Step 4: Balance the charges

1.6-Electrochemistry-Step-4-Writing-overall-redox-reactions

Step 5: Finally, balance the atoms

1.6-Electrochemistry-Step-5-Writing-overall-redox-reactions

Metals & Non-metals

Metals 

  • Metals, in general, will form positive ions by losing electrons
  • Therefore, they are oxidised and the oxidation number increases
  • Example 1:
    • When sodium reacts with water, sodium hydroxide and hydrogen gas is formed

2Na (s) + H2O (l) → 2NaOH (aq) + H2 (g)

  • The oxidation number of sodium changes from 0 to +1
  • Example 2:
    • When magnesium reacts with hydrochloric acid, magnesium chloride and hydrogen gas is formed

Mg (s) + 2HCl (l) → MgCl2 (aq) + H2 (g)

  • The oxidation number of magnesium changed from 0 to +2

Non-metals 

  • Non-metals, in general, will form negative ions by gaining electrons
  • Therefore, they are reduced and the oxidation number decreases
  • Example:
    • When sodium reacts with oxygen, sodium oxide is formed

4Na (s) + O2 (g) → Na2O (s)

  • The oxidation number of oxygen changes from 0 to -2

 

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