Redox reactions

Your two half equations are:

ClO- + 2H+ + 2e- -> Cl- + H2O
AND
ClO- + 2H2O -> ClO3- + 4H+ + 4e-

#1 x2 gets rid of the e- problem

Ahh that makes sense thank you! If I were to do that could I also cancel out the water - as there will be 2H2O on both the reactants' and products' side?

Redox is a mechanical process (working it out, I mean). I.e. follow the steps, follow them in succession and apply them, and you will get the right result

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS

I see, thanks so much for your help

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS

That looks like you're saying that if you add two half equations: one with water on the LHS and one with water on the RHS, then you'd add the waters...

Which is nonsense. You cancel the waters.

In the OP's case, there are two waters on both side (once you've multiplied to cancel the e-). The overall redox equation has no water in it. NOT four.

Redox is a mechanical process (working it out, I mean). I.e. follow the steps, follow them in succession and apply them, and you will get the right result

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS

Ahh that makes sense thank you! If I were to do that could I also cancel out the water - as there will be 2H2O on both the reactants' and products' side?

No it wouldn't. You simplify the balanced equations, so if your combined equation has an equal amount of water on the reactants and the products, you simplify it down and remove the waters, not just add the products to the reactants or vice versa.

Take the oxidation of ethanol with dichromate:

two half equations:

CH₃CH₂OH + H₂O -> CH₃COOH + 4H⁺ + 4e^-
Cr₂O₇^2- + 14H⁺ + 6e^- -> 2Cr³⁺ + 7H₂O

Multiplication is three of the first, two of the second so the combined equation is:

3CH₃CH₂OH + 3H₂O + 2Cr₂O₇^2- + 28H⁺ -> 3CH₃COOH + 12H⁺ + 4Cr³⁺ + 14H₂O

You don't then add the 3 H₂O of the left to the 14 of the right, you take them away to give

3CH₃CH₂OH + 2Cr₂O₇^2- + 16H⁺ -> 3CH₃COOH + 4Cr³⁺ + 11H₂O



Yes you could. You simplify the reaction as much as possible, so if there's 2H₂O on both sides you remove them from both sides, if there's 3H₂O in the reactants and only 2H₂O in the products then you have just the one H₂O in the reactants.