Question

In acidic solution, the nitrate ion can be used to react with a number of metal ions. One such reaction is NO3−(aq)+Sn2+(aq)→NO2(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: NO3−(aq)+Sn2+(aq)+ −−−→NO2(aq)+Sn4+(aq)+ −−− Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Answer 1

Answer:

$\boxed{\text{2, 1, 4, 2, 1, 2}}$

Explanation:

NO₃⁻ + Sn²⁺ + __ → NO₂ + Sn⁴⁺ + __

Step 1: Separate into two half-reactions.

NO₃⁻ ⟶ NO₂

Sn²⁺ ⟶ Sn⁴⁺

Step 2: Balance all atoms other than H and O.

Done

Step 3: Balance O.

NO₃⁻ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺

Step 4: Balance H

NO₃⁻ + 2H⁺ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺

Step 5: Balance charge.

NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻

Step 6: Equalize electrons transferred.

2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]

1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]

Step 7: Add the two half-reactions.

2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]

1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]                                          

     2NO₃⁻ + Sn²⁺ + 4H⁺ ⟶ 2NO₂ + Sn⁴⁺ + 2H₂O

Step 8: Check mass balance.

 On the left: 2 N, 6 O, 1 Sn, 4H

On the right: 2 N, 6 O, 1 Sn, 4H

Step 9: Check charge balance.

 On the left: -2 + 6 = +4

On the right: +4

The equation is balanced.

$\text{The coefficients are }\boxed{\textbf{2, 1, 4, 2, 1, 2}}$