[Trueolive]

Question:
Cu²⁺  CuI, I^-  I₂
So I assume I am supposed to write the two half equations and combine them?

My working:
Cu²⁺ + I^- + e^-  CuI (half equation 1)

2I^-  I₂ + 2e^- (half equation 2)

Balance electrons and thus equation ( multiply moles of equation 1 by 2)

2Cu²⁺ + 2I^- + 2e^-  2CuI

2I^-  I₂ + 2e^-

I proceeded to add the two half equations and cancelled out electrons

Answer:
2Cu²⁺ + 4I^-  I₂ + 2CuI

I am wondering if this is the correct way of getting the overall equation and if this is the correct answer.

[Borek]

Looks OK, there is just a typo in capitalization.

[Trueolive]

Thank you! Just needed to make sure this was right before redox titration calculations. (and thanks for pointing out the mistake)
 

[mjc123]

Be careful in applying your final equation; it has four I^- on the LHS, but only two of them are oxidised. (You did the calculation right in your other thread though.)

[Trueolive]

Be careful in applying your final equation; it has four I^- on the LHS, but only two of them are oxidised. (You did the calculation right in your other thread though.)

For some reason this redox equation, in particular, confused me compared to others where you balance with water, H⁺ and e^-

Like this:
IO_3-  I₂, I^-  I₂

2IO₃^- + 12H⁺ + 10e^-  I₂ + 6H₂O
2I^-  2e^- + I₂

IO₃^- + 5I^- + 6H⁺  3I₂ + 3H₂O (overall)