Textbook Question 6K.3 d)

[samanthaywu]

Can someone help me with this? I'm confused about what the reduction half reaction is because there is only one reactant.

6K.3 Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction.
d) Reaction of chlorine in water: Cl2(g) --> HClO(aq) + Cl2(g)

[Lucy Wang 2J]

2 H2O + Cl2 --> 2 HOCl + 2 H+ + 2e-
2 H2O + 2Cl2 + 2e- --> 2HOCl + 2H+ + 2Cl- + 2e-

so Cl2 is both the oxidizing and the reducing agent

[Aditya Sundaram 3D]

Can someone help me with this? I'm confused about what the reduction half reaction is because there is only one reactant.

6K.3 Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction.
d) Reaction of chlorine in water: Cl2(g) --> HClO(aq) + Cl2(g)

I think they tell us somewhere that reactants/products can be a part of both half-reactions.

[samanthaywu]

I'm still confused how you got the reduction half reaction, like what is being reduced? And also if there are 2e- on both sides of the reduction half reaction, how does that add to the oxidation half reaction to cancel out the electrons? And where did the Cl- ions come from?

2 H2O + Cl2 --> 2 HOCl + 2 H+ + 2e-
2 H2O + 2Cl2 + 2e- --> 2HOCl + 2H+ + 2Cl- + 2e-

so Cl2 is both the oxidizing and the reducing agent

[Alisa Nagashima 1B]

Hi,
There was an error in the textbook so the correct skeletal equation should be Cl₂ (g) -> HClO (aq) + Cl^- (aq)
hope this helps!