Question

Write a balanced chemical equation describing the oxidation of:

iodine solid by the manganese (III) to form the iodate ion and manganese (II) in an acidic aqueous solution. Use the smallest whole number coefficients possible. How many electrons were transferred in this redox reaction?

chloride ion by solid xenon trioxide to produce the chlorine gas and xenon gas in an acidic aqueous solution. How many electrons were transferred?

Answer 1

split in half cells

I2 = IO3-

Mn+3 = Mn+2

note that the only way there will be Mn+3 is via = HMnO4−

I2 = IO3-

HMnO4- = Mn+2

balance I

I2 = 2IO3-

H+ + MnO4-2 = Mn+2

balance O

6H2O + I2 = 2IO3-

H+ + MnO4-2 = Mn+2 + 4H2O

balance H

6H2O + I2 = 2IO3- + 12H+

7H+ + H+ + MnO4-2 = Mn+2 + 4H2O

balance charges

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

4e- + 8H+ + MnO4-2 = Mn+2 + 4H2O

balance e-

12H2O + 2I2 = 4IO3- + 24H+ + 20e-

20e- + 40H+ + 5MnO4-2 = 5Mn+2 + 20H2O

add all

20e- + 40H+ + 5MnO4-2 + 12H2O + 2I2 = 4IO3- + 24H+ + 20e- +  5Mn+2 + 20H2O

add all

16H+ + 5MnO4-2 + 2I2 = 4IO3- +  5Mn+2 + 8H2O