Question

If you complete and balance the following oxidation-reduction reaction in basic solutionNO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)how many hydroxide ions are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)?

Answer 1

Answer:

There is one 1 hydroxide ion in the balanced equation.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$NO_2^-(aq.)+Al(s)\rightarrow NH_3(aq.)+Al(OH)_4^{-}(aq)$

The half cell reactions for the above reaction follows:

Reduction half reaction:  $NO_2^-+5H_2O+6e^-\rightarrow NH_3+7OH^-$

Oxidation half reaction:  $Al+4OH^-\rightarrow Al(OH)_4^{-}+3e^-$

Multiplying the Oxidation half reaction by 2 and we get that:-

$2Al+8OH^-\rightarrow 2Al(OH)_4^{-}+6e^-$

Adding the half reactions, we get that:-

$NO_2^-+5H_2O+2Al+OH^-\rightarrow 2Al(OH)_4^{-}+NH_3$

There is one 1 hydroxide ion in the balanced equation.