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### Topic: Redox equations  (Read 575 times)

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#### Trueolive

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« on: February 11, 2022, 10:15:14 AM »
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

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 ##### Re: Redox equations
« Reply #1 on: February 11, 2022, 10:18:36 AM »
Looks OK, there is just a typo in capitalization.
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#### Trueolive

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« Reply #2 on: February 11, 2022, 10:22:36 AM »
Thank you! Just needed to make sure this was right before redox titration calculations. (and thanks for pointing out the mistake)

#### mjc123

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« Reply #3 on: February 11, 2022, 03:16:27 PM »
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

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« Reply #4 on: February 11, 2022, 05:20:41 PM »
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:
IO3- I2, I- I2

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

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