Question

Using the redox reaction below determine which element is oxidized and which is reduced. 4nh3 + 3ca(clo)2 → 2n2 + 6h2o + 3cacl2 n is oxidized and cl is reduced cl is oxidized and o is reduced cl is oxidized and n is reduced h is oxidized and n is reduced n is oxidized and o is reduced

Answer 1

The correct statement for the redox reaction $4\text{NH}_{3}+3\text{Ca(ClO)}_{2}\rightarrow2\text{N}_{2}+6\text{H}_{2}\text{O}+3\text{CaCl}_{2}$ is $\boxed{\text{N is oxidized and Cl is reduced}}$.

Further Explanation:

Redox reaction:

Redox is a term that is used collectively for the reduction-oxidation reaction. It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out simultaneously. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of gain of electrons or the decrease in the oxidation state of the atom is called reduction while that of loss of electrons or the increase in the oxidation number is known as oxidation. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as oxidant or oxidizing agent. The presence of a redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

$\text{X}+\text{Y}\rightarrow\text{X}^{+}+\text{Y}^{-}$

The oxidation half-reaction can be written as:

$\text{X}\rightarrow\text{X}^{+}+e^{-}$

The reduction half-reaction can be written as:

$\text{Y}+e^{-}\rightarrow\text{Y}^{-}$

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence,. X acts as the reducing agent whereas Y is an oxidizing agent.

Rules to calculate the oxidation states of elements:

1. The oxidation state of an element in free form is always considered zero.

2. The oxidation state of oxygen is generally taken as -2, except for peroxides.

3. The oxidation state of hydrogen is normally taken as +1.

4. The sum of oxidation states of all the elements present in a neutral compound is zero.

5. The oxidation states of group 1 and group 2 elements are +1 and +2 respectively.

The given reaction is as follows:

$4\text{NH}_{3}+3\text{Ca(ClO)}_{2}\rightarrow2\text{N}_{2}+6\text{H}_{2}\text{O}+3\text{CaCl}_{2}$

The oxidation state of H is +1.

The expression to calculate the oxidation state in $\text{NH}_{3}$ is,

$\left[(\text{oxidation state of N})+3(\text{oxidation state of H})\right]=0$                 ...... (1)

Rearrange equation (1) for the oxidation state of N.

$\text{oxidation state of N}=\left[0-3(\text{oxidation state of H})\right]$                 ...... (2)

Substitute +1 for the oxidation state of H in equation (2).

$\begin{aligned}\text{oxidation state of N}&=\left[0-3(+1)\right]\\&=[0-3]\\&=-3\end{aligned}$

As the oxidation state of the free element is zero. $\text{N}_{2}$ is also a free element so its oxidation state is also 0.

The oxidation state of N increases from -3 to 0. So it is being oxidized in the reaction.

The oxidation state of O is -2 and that of Ca is +2.

The expression to calculate the oxidation state in $\text{Ca(ClO)}_{2}$ is,

$\left[(\text{oxidation state of Ca})+2(\text{oxidation state of Cl})+2(\text{oxidation state of O})\right]=0$                 ...... (3)

Rearrange equation (3) for the oxidation state of Cl.

$\text{oxidation state of Cl}=\dfrac{[-(\text{oxidation state of Ca})-2(\text{oxidation state of O})]}{2}$                ...... (4)

Substitute -2 for the oxidation state of O and +2 for the oxidation state of Ca in equation (4).

$\begin{aligned}\text{oxidation state of Cl}&=\dfrac{[-(+2)-2(-2)]}{2}\\&=\dfrac{[-2+4]}{2}\\&=+1\end{aligned}$

The oxidation state of Ca is +2.

The expression to calculate the oxidation state in $\text{CaCl}_{2}$ is,

$\left[(\text{oxidation state of Ca})+2(\text{oxidation state of Cl})\right]=0$                             ...... (5)

Rearrange equation (5) for the oxidation state of Cl.

$\text{oxidation state of Cl}=\dfrac{[0-(\text{oxidation state of Ca})]}{2}$                          ...... (6)

Substitute +2 for the oxidation state of Ca in equation (6).

$\begin{aligned}\text{oxidation state of Cl}&=\dfrac{[0-(+2)]}{2}\\&=\dfrac{[0-2]}{2}\\&=-1\end{aligned}$

The oxidation state of Cl decreases from +1 to -1. So it is being reduced.

So the first option is correct that N is oxidized and Cl is reduced.

Learn more:

1. Which of the following occur during redox reaction: brainly.com/question/1616320

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords:

redox reaction, CaCl2, NH3, N2, Ca(ClO)2, Cl, Ca, O, oxidized, reduced, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously.

Answer 2

The oxidation number of elements in equation below are,

                         4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂

O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1

Oxidation:
               Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).

Reduction:
               Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).

Result:
          N is oxidized and Cl is reduced.