IO3^- ==> I2
First balance the I so we are working with apples and apples and not apples and oranges.
I on left is +10; on right is 0
add 10 e to left side
2IO3^- + 10e ==> I2
charge on left is -12; right is zero.
Add H^+ to left and H2O to the right.
2IO3^- + 10e + 12H^+ ==> I2 + 6H2O
I^- ==> I2
balanced I (again for apples and apples)
2I^- ==>I2 + 2e
Now multiply 1st half equation by 1 and second by 5 and add them.
When finished you will find that you can divide all of the coefficients by 2.
Balance Redox using reactions by the 1/2 reaction method.
NaIO3 (aq) + KI (aq) -->„³ I2 (s)
2 answers
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