Cu^2+ + I^- --> CuI + I3^-
I'm not sure how to divide this into two separate reactions?
Please explain.
I'm not sure how to divide this into two separate reactions?
Please explain.
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It's tricky.
Try the following:
Cu^2+ + I^- + e^- --> CuI
3 I^- --> I3^- + 2 e^- >>> Start with 2 I^- --> I2 + 2 e^- then add I^- to react with I2 producing the more stable I3^-. Now, multiply the 1st reaction by 2 to obtain the correct numbers of electrons lost and gained. Then add the two half-reactions algebraically.
2 Cu^2+ + 2 I^- + 2 e^- --> 2 CuI
3 I^- --> I3^- + 2 e^-
-------------- ------------------- -----------
2 Cu^2+ + 5 I^- --> 2 CuI + I3^-
That's how you do it.
Keep on Truckin!!
Try the following:
Cu^2+ + I^- + e^- --> CuI
3 I^- --> I3^- + 2 e^- >>> Start with 2 I^- --> I2 + 2 e^- then add I^- to react with I2 producing the more stable I3^-. Now, multiply the 1st reaction by 2 to obtain the correct numbers of electrons lost and gained. Then add the two half-reactions algebraically.
2 Cu^2+ + 2 I^- + 2 e^- --> 2 CuI
3 I^- --> I3^- + 2 e^-
-------------- ------------------- -----------
2 Cu^2+ + 5 I^- --> 2 CuI + I3^-
That's how you do it.
Keep on Truckin!!