Question

What is the half equation of the reduction reaction:
Cl2(g)+H2O(l) -> HOCl(aq) + H+(aq) + Cl- (aq)

Answer 1

You have to find the oxidation states of each species in each formula, then compare the products to the reactants and find out which species gains electrons. To start, you have\[Cl_2 + H_2O \rightarrow HOCl + H^{+1} + Cl^{-1}\] If we add the oxidation states to each species in the reaction, we get\[Cl_2^{(0)} + H^{(+1)}_2O^{(-2)} \rightarrow H^{(+1)}O^{(-2)}Cl^{(+1)} + H^{+1} + Cl^{-1}\]so we can see that the hydrogens and oxygens DON'T CHANGE in their oxidation states. The chlorine starts as a 0, then one Cl becoems a +1 forming HOCl, and the other becomes a -1 forming the Cl-1 ion. Which one of these is reduction?
Gaining an electron is reduction, so the reduction half-reaction is\[Cl_2 + 2e^{-1} \rightarrow 2Cl^{-1}\] balanced for mass and charge, of course

Answer 2

How can it be 2Cl- if one of them became a Cl+?

Answer 3

If we left it as\[Cl_2 + 1e^{-1} \rightarrow Cl^{-1}\] then mass wouldn't balance. We'd have 2 Cl's on one side, but only 1 Cl on the other. Each half reaction must balance with itself first, then we can take 2 balanced half-reactions and put them back together, and know that they will still be balanced. The other half-reaction will be\[2H_2O + Cl_2 \rightarrow 2HOCl + 2H^{+1} + 2e^{-1}\] when we pair that up with the original, balanced, reduction half-reaction\[Cl_2 + 2e^{-1} \rightarrow 2Cl^{-1}\] and add those two back together again, we still get a reaction that is balanced for both mass and charge

Answer 4

sorry not getting it,
where do the two electrons from the cl go if one ends up as cl+?
thanks,

Answer 5

the 2 electrons that one Cl loses go into the other Cl ion, that's what makes it a redox reaction.

Answer 6

Got it, thanks!
Sorry it's taking me a while to get back to you.