in studying the Cr2O72-/Cr3+ and NO3-/HNO2 oxidation/ reduction equilibrium reac

Question

in studying the Cr2O72-/Cr3+ and NO3-/HNO2 oxidation/ reduction equilibrium reaction in acidic solution, a student finds out that HNO2 is oxidized to nitrate ion, NO3-, by dichromate ion, Cr2O72- while, the dichromate ion is reduced to chromium (III) ion, Cr3+ (Note: electrons "e-" are used to balance the ions charges in the half reactions and should cancel out in the "NET" redox reaction of reactants verus products)

(a) Balance the following half reaction (Cr2O72- is reduced to Cr3+)

H+ + Cr2O72- + e (arrow) Cr3+ + H2O

(b) Balance the half oxidation reaction (HNO2 is oxidized to NO3-)

HNO2 + H2O (arrow) NO3- + H+ + e-

(c) What is the "NET" redox reaction combining the two half reactions above in questions (a) and (b) and balancing charges?

Explanation / Answer

(a) H+ + Cr2O7^2- + e- ---> Cr3+ + H2O

Cr is going from +6 to +3

14H+ + Cr2O7^2- + 3e- ---> 2Cr3+ + 7H2O

Is the balanced equation

(b) HNO2 + H2O ---> NO3- + H+ + e-

N is going from +3 to +5 oxidation state

HNO2 + H2O ---> NO3- + 3H+ + 2e-

Is the balanced equation

(c) Net redox reaction,

(14H+ + Cr2O7^2- + 3e- ---> 2Cr3+ + 7H2O) x 2

(HNO2 + H2O ---> NO3- + 3H+ + 2e-) x 3

-----------------------------------------------------------------

2Cr2O7^2- + 3HNO2 + 19H+ ---> 4Cr3+ + 3NO3- + 11H2O

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