Chemistry alevel aqa

Write the 2 ionic half equations form the following full ionic equation:
2IO3−+5HSO3−→3HSO4−+2SO42−+I2+H2O
can someone explain this
thank you

Reply 1

TypicalNerd

Original post

by LisaPerera

Write the 2 ionic half equations form the following full ionic equation:
2IO3−+5HSO3−→3HSO4−+2SO42−+I2+H2O
can someone explain this
thank you

Let’s begin by identifying which elements change oxidation states. An easy place to start is to find an element in its elemental state (e.g. I2, S, Fe etc) in the equation as this is guaranteed to have been oxidised or reduced.

We can see IO3^- becomes I2, so iodine changes oxidation state from +5 (since O is virtually always -2) to 0. This means it is reduced (since the oxidation state decreases) and so it gains 5 electrons (since the change in oxidation states is 5).

Hence, we have the unbalanced

IO3^- + 5e^- —> I2

Obviously we can balance the iodines by scaling one side of the equation as appropriate, but it’s less clear what to do about the oxygens and the charges. Fortunately, the equation does give us a crucial hint - water is produced, so that’s how the oxygens are lost. This incidentally means we need H^+ on the LHS as well. Let’s start in the following logical order:

Scaling up the LHS by 2 to balance the iodines:

2IO3^- + 10e^- —> I2

Adding H2O to the RHS and H^+ to the RHS:

2IO3^- + H^+ + 10e^- —> I2 + H2O

Realising to balance the oxygens, there must be 6H2O on the RHS:

2IO3^- + H^+ + 10e^- —> I2 + 6H2O

Realising to balance the hydrogens (and the charges), there must be 12H^+ on the LHS:

2IO3^- + 12H^+ + 10e^- —> I2 + 6H2O

And that’s the reduction half-equation derived. Could you try using a similar approach to work out the other half-equation?