If you complete and balance the following oxidation-reduction reaction in basic solution no2−(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)? if you complete and balance the following oxidation-reduction reaction in basic solution how many hydroxide ions are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)? one on the reactant side one on the product side four on the reactant side seven on the product side none

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

Step 2: Balancing electrons

NO2- → NH3

On the left side N has an oxidation number of +3 and on the right side -3.

NO2- +6e-→ NH3

Al(s) → Al(OH)4-

On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

NO2- +6e-→ NH3

2(Al(s) → Al(OH)4- +6e-)

Step 3: Balance with OH/H2O

NO2- +6e +5H2O → NH3 +7OH-

2Al +8OH- → 2Al(OH)4- + 6e-

Step 4: The netto reaction

NO2- + 5H2O + 2Al + 8OH-  → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

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