I-(aq) + IO3-(aq) → I2(aq) (acidic solution)?

Question

How do I make this finished: I-(aq) + IO3-(aq) → I2(aq) (acidic solution)

Answer 1

Balance the two half-reactions:

2I- ===> I2 + 2e-

12H+ + 2IO3- + 10e- ===> I2 + 6H2O

Multiply the first half-reaction by 2 to make 10e- lost to match the 10e- gained in the second equation, and add them together.

12H+ + 10I- + 2IO3- ===> 6I2 + 6H2O

6H+ + 5I- + IO3- ===> 3I2 + 3H2O

Answer 2

I- and IO3- gives I2
We divide the reaction in two half-reactions
(2 I- >> I2 + 2e-) x5 ( oxidation : I goes from -1 to 0 )
2 IO3- + 12H+ + 10e- >> I2 + 6H2O ( reduction : I goes from +5 to 0 )

10 I- >> 5I2 + 10e-
2IO3- + 12H+ + 10e- >> I2 + 6H2O
-----------------------------------------------------

10 I- + 2IO3- + 12H+ >> 6I2 + 6H2O
To get the smallest numbers we divide by 2 :
5 I- + IO3- + 6H+ >> 3I2 + 3H2O

Answer 3

Both the I- and the IO3- make I2 so write out the 1/2 reactions and solve from there.

2I-(aq) --> I2(aq) + 2e- * 5

2IO3-(aq) + 12H+(aq) + 10e- --> I2 + 6H2O(l)

add

10I-(aq) + 2IO3-(aq) + 12H+(aq) --> 6I2(aq) + 6H2O(l)

multiple of 2 so divide equation by 2

5I-(aq) + IO3-(aq) + 6H+(aq) --> 3I2(aq) + 3H2O(l)

combine

5HI(aq) + HIO3(aq) --> 3I2(aq) + 3H2O(l)

Answer 4

Cr2O7{2−}(aq)+6Cu{+}(aq)+14H{+}(aq)→2Cr{...