Balance the following redox reaction, then identify the oxidizing and reducing agents.?

Question

IO4- + I- + H+---> I2 + H2O (all #s denote subscripts)

Please explain how you got to each step...how you balanced it, then how you ultimately identified the oxidizing and reducing agents. Thank you.

Ok, getting the balanced equation isn't the issue: IO4- + 7I- + 8H+ --> 4I2 + 4H2O ... that can be arrived at by simple inspection, but identifying the oxidation and reduction agents are what's troubling me...

Answer 1

IO4- + 7 I- + 8 H+ --> 4 I2 + 4 H2O

Both the I in the IO4- and in the I- end up as I2. The I in IO4- decreases from +7 to 0 in I2, while the I in I- increases from -1 to 0 in I2. So the ratio of IO4- to I- is 1:7.

IO4- + 7 I- + H+ --> I2 + H2O

This give 8 atoms of I, so

IO4- + 7 I- + H+ --> 4 I2 + H2O

Now balance the charges

IO4- + 7 I- + 8 H+ --> 4 I2 + H2O

and the the H

IO4- + 7 I- + 8 H+ --> 4 I2 + 4 H2O

finally check the O (4 on each side, they are balanced).

The IO4- --> I2 and I- --> I2 are the redox, see electron count above.

The chemical way to look at redox processes is that the reductant transfers electrons to the oxidant. Thus, in the reaction, the reductant or reducing agent loses electrons and is oxidized and the oxidant or oxidizing agent gains electrons and is reduced.

I- loses electron. It is the reducing agent. The IO4- gains electrons. It is the oxidizing agent.

Answer 2

What element is being oxidized in the following redox reaction? MnO4- (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) C