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.

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.

that's just how they are

explanation:

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