Consider the reaction: Zn(s) + NO₃-(aq) $\rightarrow$ NH₃(aq) + Zn(OH)₄²^-(aq) When this equation is balanced in basic solution, the coefficient for water will be__________, and the number of electrons transferred will be __________.

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First, we balance the half-reactions: Reduction: NO₃-(aq) F1F1F1S1 + 8H2O + 3e- → NH₃(aq) + 10OH-²^-(aq) Oxidation: Zn(s) + 4OH-²^-(aq) → Zn(OH)₄²^-(aq) + 2e- Next, we balance the electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, giving us: Reduction: 2NO₃-(aq) F1F1F1S1 + 16H2O + 6e- → 2NH₃(aq) + 20OH-²^-(aq) Oxidation: 3Zn(s) + 12OH-²^-(aq) → 3Zn(OH)₄²^-(aq) + 6e- Next, we add the half-reactions: 3Zn(s) + 12OH-²^-(aq) + 2NO₃-(aq) F1F1F1S1 + 16H2O → 3Zn(OH)₄²^-(aq) + 2NH₃(aq) + 20OH-²^-(aq) We can simplify by canceling any species that appear on both sides: 3Zn + 2NO3- + 12OH- → 3Zn(OH)2 + 2NO + 6H2O + 6e- Finally, we balance the equation by adding water to the side that needs it. Since we have 12 hydroxide ions on the left side and 6 water molecules equal to 12 hydrogens and 6 oxygens, we need to add 6 more water molecules to the left side: 3Zn + 2NO3- + 12OH- + 6H2O → 3Zn(OH)2 + 2NO + 6H2O + 6e- The coefficient for water is 6, and the number of electrons transferred is 6+2(3) = 12, which is equal for both the oxidation and reduction half-reactions.

Consider the Reaction: Zn(s) + NO3-(Aq) →\rightarrow→ NH3(aq) + Zn(OH)42-(aq) When This Equation Is Balanced in Basic

Consider the Reaction: Zn(s) + NO₃-(Aq) $\rightarrow$ NH₃(aq) + Zn(OH)₄²^-(aq) When This Equation Is Balanced in Basic