These are half equations for some reactions where positive ions gain electrons:
Na+ + e- → Na
Pb2+ + 2e- → Pb
2H+ + 2e- → H2
Balance the half equation for the formation of aluminium during electrolysis: Al3+ + e- → Al.
The balanced half equation is: Al3+ + 3e- → Al (because three negatively charged electrons are needed to balance the three positive charges on the aluminium ion).
Write a balanced half equation for the formation of calcium from a calcium ion, Ca2+.
Ca2+ + 2e- → Ca
These are half equations for some reactions where negatively charged ions lose electrons:
2Cl- → Cl2 + 2e-
2O2- → O2 + 4e-
Write a balanced half equation for the formation of bromine, Br2, from bromide ions, Br-.
2Br- → Br2 + 2e-
For example, silver nitrate solution reacts with sodium chloride solution. The products are insoluble solid silver chloride and sodium nitrate solution:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms needs to be shown:
Ag+(aq) + Cl-(aq) → AgCl(s)
In a balanced ionic equation:
Explain why this ionic equation is balanced:
Ba2+(aq) + SO42-(aq) → BaSO4(s)
There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).
Balance this ionic equation, which represents the formation of a silver carbonate precipitate:
Ag+(aq) + CO32-(aq) → Ag2CO3(s)
2Ag+(aq) + CO32-(aq) → Ag2CO3(s)
Balance this ionic equation, which represents the formation of an aluminium hydroxide precipitate:
Al3+(aq) + ...OH- (aq) → Al(OH)3(s)
Al3+(aq) + 3OH- (aq) → Al(OH)3(s)