Question

Balance the following skeleton reaction in acidic solution (if the coefficient is 1, put 1; if the coefficient is 0, put 0) and identify the oxidizing and reducing agents. For the oxidizing and reducing agent, enter numbers and +/- directly for superscripts and subscripts. (eg. enter HCO3- for HCO3-, Cu2+ for Cu2+) CrO42- (aq) + N2O (g) Cr3+ (aq) + NO (g) 2 CrO42–(aq) + 3 N2O(g) + 10 H+(aq) + 0 H2O(l) 2 Cr3+(aq) + 6 NO(g) + 0 H+(aq) + 5 H2O(l) (acidic) Oxidizing agent: CrO42- Reducing agent: N2O

Answer 1

Answer:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Explanation:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Oxidizing agent: -----------------------------> CrO42-

Reducing agent: ----------------------------> N2O

explanation:

in CrO4-2 oxdiation state of Cr = +6

in Cr+3 oxidation state = +3

+6 oxidation state changed from +3 it is reduction .

so CrO4-2 is oxidizing agent

atomatically

N2O should be reducing agent