Respuesta :
Answer:
[tex]4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)[/tex]
-) Oxidizing agent: [tex]NO_3^-_(_a_q_)[/tex]
-) Reducing agent: [tex]Zn_(_s_)[/tex]
Explanation:
The first step is separate the reaction into the semireactions:
A.[tex]Zn~->Zn(OH)_4^-^2[/tex]
B.[tex]NO_3^-~->~NH_3[/tex]
If we want to balance in basic medium we have to follow the rules:
1. We adjust the oxygen with [tex]OH^-[/tex]
2. We adjust the H with [tex]H_2O[/tex]
3. We adjust the charge with [tex]e^-[/tex]
Lets balance the first semireaction A. :
[tex]Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-[/tex]
Now, lets balance semireaction B:
[tex]NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-[/tex]
Finally, we have to add the two semireactions:
_________________________________________
[tex]8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)[/tex]
[tex]2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)[/tex]
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[tex](8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)[/tex]
[tex](2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)[/tex]
Cancel out the species on both sides:
[tex]8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3[/tex]
Simplifying the equation :
[tex]4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3[/tex]
The [tex]Zn_(_s_)[/tex] is oxidized therefefore is the reducing agent. The [tex]NO_3^-_(_a_q_)[/tex]is reduced therefore is the oxidizing agent.
