Write balanced net ionic equations for the following reactions in basic solution.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Zn(s)+NO3-(aq)--->NH3(aq)+Zn(OH)4(2-)(…
Express your answer as a chemical equation. Identify all of the phases in your answer.
Zn(s)+NO3-(aq)--->NH3(aq)+Zn(OH)4(2-)(…
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This reaction is a redox reaction under basic conditions. Balance the equation by using the half-reaction method of balancing redox equations.
Oxidation half-reaction:
1st balance the mass on each side:
4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq)
Now balance the charge on each side by adding electrons to the side that is deficient in negative charge.
4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq) + 2 e^- Note: Charge is -4 on both sides. This is balanced.
Reduction half-reaction:
Follow the same procedure as before. This is more difficult because you need to first balance it as though in acid conditions then add as many OH^- ions to each side of the half-reaction as there are H^+ ions in the equation.
1st balance for acid: 9 H^+(aq) + NO3^-(aq) --> NH3(aq) + 3 H2O(l)
Next add OH^-
9 OH^-(aq) + 9 H^+(aq) + NO3^-(aq) --> NH3 + 3 H2O(l) + 9 OH^-
Next combine H^+ and OH^- that are on the same side of the arrow.
9 H2O(l) + NO3^-(aq) --> NH3(aq) + 3 H2O(l) + 9 OH^-
Subtract 3 H2O from each side:
6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^-
Now, balance the charge by adding electrons to the left side.
8 e^- + 6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^- Note: Charge balanced on each side.
When you balance redox reactions the electrons lost must equal electrons gained. In the oxidation half-reaction there are only 2 electrons so the entire equation must be multiplied by 4. Then the two equations will be added to each other for the final net ionic equation.
4(4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq) + 2 e^-)
16 OH^-(aq) + 4 Zn(s) --> 4 Zn(OH)4^2-(aq) + 8 e^-
8 e^- + 6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^-
Add the last two algebraically, subtracting out the things that are the same on both sides.
4 Zn(s) + 7 OH^-(aq) + 6 H2O(l) + NO3^- --> 4 Zn(OH)4^2-(aq) + NH3(aq)
Hope this is helpful to you. JIL HIR
Oxidation half-reaction:
1st balance the mass on each side:
4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq)
Now balance the charge on each side by adding electrons to the side that is deficient in negative charge.
4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq) + 2 e^- Note: Charge is -4 on both sides. This is balanced.
Reduction half-reaction:
Follow the same procedure as before. This is more difficult because you need to first balance it as though in acid conditions then add as many OH^- ions to each side of the half-reaction as there are H^+ ions in the equation.
1st balance for acid: 9 H^+(aq) + NO3^-(aq) --> NH3(aq) + 3 H2O(l)
Next add OH^-
9 OH^-(aq) + 9 H^+(aq) + NO3^-(aq) --> NH3 + 3 H2O(l) + 9 OH^-
Next combine H^+ and OH^- that are on the same side of the arrow.
9 H2O(l) + NO3^-(aq) --> NH3(aq) + 3 H2O(l) + 9 OH^-
Subtract 3 H2O from each side:
6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^-
Now, balance the charge by adding electrons to the left side.
8 e^- + 6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^- Note: Charge balanced on each side.
When you balance redox reactions the electrons lost must equal electrons gained. In the oxidation half-reaction there are only 2 electrons so the entire equation must be multiplied by 4. Then the two equations will be added to each other for the final net ionic equation.
4(4 OH^-(aq) + Zn(s) --> Zn(OH)4^2-(aq) + 2 e^-)
16 OH^-(aq) + 4 Zn(s) --> 4 Zn(OH)4^2-(aq) + 8 e^-
8 e^- + 6 H2O(l) + NO3^-(aq) --> NH3(aq) + 9 OH^-
Add the last two algebraically, subtracting out the things that are the same on both sides.
4 Zn(s) + 7 OH^-(aq) + 6 H2O(l) + NO3^- --> 4 Zn(OH)4^2-(aq) + NH3(aq)
Hope this is helpful to you. JIL HIR
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Zn (0) -2 e >>> Zn (2++
N (5+) +2e >>> N (3+)
add H2O on the left and OH- on the right
N (5+) +2e >>> N (3+)
add H2O on the left and OH- on the right