For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this reaction in basic solution. The phases are optional. balanced reaction: NO-2 + Ag^{+} -> NO-3+ Ag

The following are the steps  to complete and balnce the equation for the given reaction

Explanation:

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ > NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ > NO3– + Ag(s)

N= +3                           N = +5                        

O = -2                            O = -2

Ag = +1                         Ag = 0

NO2– > NO3– ………oxidation half reaction

Ag+ > Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

     NO2– > NO3–

     Ag+ > Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

     NO2– + H2O > NO3–

     Ag+ > Ag(s)

Step 4) Balance the H by adding H+

    NO2– + H2O > NO3– + 2H+

     Ag+ > Ag(s)

Step 5) Balance the charge by adding electron

    NO2– + H2O > NO3– + 2H+ + 2e-

     Ag+ + 1e> Ag(s)

Step 6) Balance the electron in both half reaction

    NO2– + H2O > NO3– + 2H+ + 2e-

     2 Ag+ + 2e> 2 Ag(s)

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a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o