Write a balanced half-reaction for the oxidation of gaseous nitric oxide (NO) to aqueous nitrous acid (HNO2) in acidic aqueous solution.

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

Step-by-step explanation:

NO ⇒ oxidation number of N = +2

HNO₂ ⇒ oxidation number of N= +3

Therefore, NO has to lose 1 electron to be oxidized to HNO₂. We write the half-reaction with 1 electron (1 e-) in the products side.

NO → HNO₂ + 1 e-

Now, we have 0 electrical charges in the reactants side, and a total of -1 electrical charge in the products side. As the reaction is in acidic aqueous solution, we have to add H⁺ ions to balance the charges. We perform the balance by adding 1 H⁺ (positive charge) to neutralize the negative charge in the side of the products:

NO → HNO₂ + 1 e- + 1 H⁺

Now, we perform the mass balance. We have:

N: 1 atom in both sides

O: 1 atom in reactants side and 2 atoms in products side

H: 0 atoms in reactants side, 2 atoms in products side.

Thus, we have to add 1 H₂O molecule to the reactants side to equal the masses:

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

Finally, the oxidation half-reaction is:

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml