What are the oxidation number of nitrogen in the reactants and products of this reaction? NH4+(aq)+2H2O(l) —> 4H+(aq)+NO2^-(aq)+6e^- Find NH4^+ and NO2^-Then answer…..A voltaic cell with an alkaline aqueous electrolyte is based on the reaction between MnO4^-(aq) and Cd(s), producing MnO2(s) and Cd^2+(aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.
The correct answer and explanation is :
Let’s break this into two parts:
1. Oxidation Numbers in the Reaction:
Given the reaction:
[ \text{NH}_4^+(aq) + 2\text{H}_2\text{O}(l) \rightarrow 4\text{H}^+(aq) + \text{NO}_2^-(aq) + 6e^- ]
We are tasked with finding the oxidation numbers of nitrogen in both the reactants (NH₄⁺) and products (NO₂⁻).
In NH₄⁺ (Ammonium ion):
- The ammonia molecule, NH₃, has nitrogen with an oxidation number of -3. In the ammonium ion (NH₄⁺), the overall charge is +1.
- The four hydrogen atoms each have an oxidation number of +1.
So, the equation for the oxidation number of nitrogen in NH₄⁺ is:
[
\text{Oxidation number of N} + 4(\text{Oxidation number of H}) = +1
]
[
\text{Oxidation number of N} + 4(1) = +1
]
[
\text{Oxidation number of N} = +1 – 4 = -3
]
Thus, the oxidation number of nitrogen in NH₄⁺ is -3.
In NO₂⁻ (Nitrite ion):
- The nitrite ion (NO₂⁻) has a total charge of -1.
- Oxygen has an oxidation number of -2. There are two oxygen atoms in the ion.
We can set up the equation for nitrogen’s oxidation number:
[
\text{Oxidation number of N} + 2(\text{Oxidation number of O}) = -1
]
[
\text{Oxidation number of N} + 2(-2) = -1
]
[
\text{Oxidation number of N} – 4 = -1
]
[
\text{Oxidation number of N} = -1 + 4 = +3
]
Thus, the oxidation number of nitrogen in NO₂⁻ is +3.
Summary:
- In NH₄⁺, the oxidation number of nitrogen is -3.
- In NO₂⁻, the oxidation number of nitrogen is +3.
2. Voltaic Cell Based on MnO₄⁻ and Cd(s):
In this voltaic cell, the reaction is between the permanganate ion (MnO₄⁻) and solid cadmium (Cd), producing manganese dioxide (MnO₂) and cadmium ions (Cd²⁺).
Half-Reactions:
- Anode (oxidation):
The cadmium metal is oxidized at the anode. The half-reaction is:
[
\text{Cd}(s) \rightarrow \text{Cd}^{2+}(aq) + 2e^-
] Cadmium (Cd) loses two electrons to form Cd²⁺. The oxidation number of cadmium increases from 0 to +2. - Cathode (reduction):
The permanganate ion (MnO₄⁻) is reduced at the cathode. The half-reaction is:
[
\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{MnO}_2(s) + 4\text{H}_2\text{O}(l)
] Here, the MnO₄⁻ ion gains electrons (reduction) to form MnO₂, where the oxidation state of manganese decreases from +7 to +4.
Balanced Cell Reaction:
To balance the cell reaction, we need to balance the number of electrons involved in each half-reaction. The anode half-reaction involves 2 electrons, while the cathode half-reaction involves 5 electrons. To equalize the electrons, multiply the anode half-reaction by 5 and the cathode half-reaction by 2:
- Anode:
[
5\text{Cd}(s) \rightarrow 5\text{Cd}^{2+}(aq) + 10e^-
] - Cathode:
[
2\text{MnO}_4^-(aq) + 16\text{H}^+(aq) + 10e^- \rightarrow 2\text{MnO}_2(s) + 8\text{H}_2\text{O}(l)
]
Now add both half-reactions:
[
5\text{Cd}(s) + 2\text{MnO}_4^-(aq) + 16\text{H}^+(aq) \rightarrow 5\text{Cd}^{2+}(aq) + 2\text{MnO}_2(s) + 8\text{H}_2\text{O}(l)
]
This is the overall balanced cell reaction.
Explanation:
- In a voltaic cell, electrons flow from the anode (oxidation site) to the cathode (reduction site) through an external circuit, generating electrical energy.
- Oxidation occurs at the anode (where cadmium loses electrons), and reduction occurs at the cathode (where MnO₄⁻ gains electrons).
- The cell reaction involves the transfer of electrons from Cd to MnO₄⁻, changing their oxidation states and forming products like MnO₂ and Cd²⁺.