Question

Consider the following redox reaction (under acidic conditions) when answering this question:

H5IO6 +NO = IO3^-1 + NO3^-1

A. Neatly write the balanced reaction for the first reactant.

B. Neatly write the balanced half reaction for the second reactant.

C. Neatly write the full balanced redox reaction. SHOW WORK.

D. Specifically, by nothing changes in the oxidation numbers, explian which atom in which species is being reduced

Answer 1

a)

BAlanced reaction

H5IO6 + NO = IO3-1 + NO3-

split

H5IO6 = IO3-

NO = NO3-

balance atoms of O adidng H2O

H5IO6 = IO3- + 3H2O

2H2O + NO = NO3-

balance H+

H+ + H5IO6 = IO3- + 3H2O

2H2O + NO = NO3- + 4H+

balance charges

4e- + H+ + H5IO6 = IO3- + 3H2O

2H2O + NO = NO3- + 4H+ + e-

balance e-

4e- + H+ + H5IO6 = IO3- + 3H2O

8H2O + 4NO =4NO3- + 16H+ + 4e-

add all

8H2O + 4NO + 4e- + H+ + H5IO6 = IO3- + 3H2O + 4NO3- + 16H+ + 4e-

cancel common terms

5H2O + 4NO + H5IO6 = IO3- + 4NO3- + 1H+

therefore:

for the first reactant, i.e. th eacid:

4e- + H+ + H5IO6 = IO3- + 3H2O

B)

for the second reactant

8H2O + 4NO =4NO3- + 16H+ + 4e-

simplest :

2H2O + NO = NO3- + 4H+ + e-

c)

redox reaction must include oxidation + reduciton

the reactions:

4e- + H+ + H5IO6 = IO3- + 3H2O

8H2O + 4NO =4NO3- + 16H+ + 4e-

5H2O + 4NO + H5IO6 = IO3- + 4NO3- + 1H+

d)

clearly --> oxidation is that which loses electrons, so

8H2O + 4NO =4NO3- + 16H+ + 4e-

meaning that NO is oxidizing, and makes sense since it gains oxygen

5H2O + 4NO + H5IO6 = IO3- + 4NO3- + 1H+

reduction is that which gains electrons, so.

4e- + H+ + H5IO6 = IO3- + 3H2O

clearly, Iodine is reducgin, from IO6 to IO3-