half-reaction method question

Question

Answer the following questions and balance the equation below using the half-reaction method

ClO₃⁻_(aq)  +  Cu(CN)₂⁻_(aq) →     Cu²⁺_(aq)  +  CN⁻_(aq)  +   Cl_2(aq)    

To balance the number of electrons, the reduction half-reaction must be multiplied by?  

Answer 1

Not exactly sure how to answer this since the question didn't mention if it was in acid or base conditions. Also, not sure that it is the reduction reaction that is multiplied to balance the electrons. But here is my attempt (using acidic conditions):

2ClO₃^- + 12H⁺ + 10e- ===> Cl₂ + 6H₂O Reduction half reaction

Cu⁺ ===>Cu²⁺ + e- Oxidation half reaction (x10)

2ClO₃^- + 12H⁺ + 10Cu⁺ ===> Cl₂ + 6H₂O + 10Cu²⁺ (CN^- is a spectator)

To balance the reaction, the OXIDATION half reaction was multiplied by 10.