Question

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents. Zn(s) + NO3−(aq) → Zn(OH)42−(aq) + NH3(g) [basic]

Answer 1

Answer:

$4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)$

-) Oxidizing agent: $NO_3^-_(_a_q_)$

-) Reducing agent: $Zn_(_s_)$

Explanation:

The first step is separate the reaction into the semireactions:

A.$Zn~->Zn(OH)_4^-^2$

B.$NO_3^-~->~NH_3$

If we want to balance in basic medium we have to follow the rules:

1. We adjust the oxygen with $OH^-$

2. We adjust the H with $H_2O$

3. We adjust the charge with $e^-$

Lets balance the first semireaction A. :

$Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-$

Now, lets balance semireaction B:

$NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-$

Finally, we have to add the two semireactions:

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$8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)$

$2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)$

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$(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)$

$(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)$

Cancel out the species on both sides:

$8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3$

Simplifying the equation :

$4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3$

The $Zn_(_s_)$ is oxidized therefefore is the reducing agent. The $NO_3^-_(_a_q_)$is reduced therefore is the oxidizing agent.