Question

Which substance is the reducing agent in this reaction? 16h++2cr2o72−+c2h5oh→4cr3++11h2o+2co2?

Answer 1

Answer is: the reducing agent is C₂H₅OH.

Balanced chemical reaction:

16H⁺ + 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺ + 11H₂O + 2CO₂.

Oxidation half reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O/ ×2.

2Cr₂O₇²⁻ + 28H⁺ + 12e⁻ → 4Cr³⁺ + 14H₂O.

Reduction half reaction: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻.

Net reaction: 2Cr₂O₇²⁻ + 28H⁺ + C₂H₅OH + 3H₂O → 4Cr³⁺ + 14H₂O + 2CO₂ + 12H⁺.

Reducing agent is element or compound who loose electrons in chemical reaction. Ethanol (C₂H₅OH) lost electrons and it is oxidized to carbon dioxide (CO₂).

Answer 2

The reducing agent in the reaction is $\boxed{\text{C}_{2}\text{H}_{5}\text{OH}}$.

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 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 number of free element is always zero.

2. The oxidation number 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 numbers of all the elements present in a neutral compound is zero.

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

The oxidation state of O is -2.

The expression to calculate the oxidation number in $\text{Cr}_{2}\text{O}_{7}^{2-}$ is:

$\left[2(\text{Oxidation state of Cr})+7(\text{oxidation state of O})\right]=-2$     ...... (1)

Rearrange equation (1) for oxidation number of Cr

$2(\text{Oxidation number of Cr})=\left[(-2)-7(\text{oxidation number of O})\right]$                  …… (2)

Substitute -2 for oxidation state of O in equation (2).

$\begin{aligned}\text{Oxidation state of Cr}&=\dfrac{[(-2)-7(-2)]}{2}\\&=\dfrac{[-2+14]}{2}\\&=+6\end{aligned}$

The charge on Cr is +3 so its oxidation state is also +3. The oxidation state of Cr changes from +6 to +3. This shows it decreases during the reaction and therefore Cr is an oxidizing agent.

The oxidation state of O is -2 and the oxidation state of H is +1.

The expression to calculate the oxidation in $\text{C}_{2}\text{H}_{5}\text{OH}$ is:

$\left[2(\text{oxidation state of C})+1(oxidation state of O)+6(oxidation state of H)\right]=0$   ...... (3)

Rearrange equation (3) for oxidation state of C

$2(\text{oxidation state of C})=\left[-1(oxidation state of O)-6(oxidation state of H)\right]$       ...... (4)

Substitute -2 for oxidation state of O and +1 for oxidation state of H in equation (4).

$\begin{aligned}\text{Oxidation state of C}&=\dfrac{[-1(-2)-6(+1)]}{2}\\&=\dfrac{[+2-6]}{2}\\&=-2\end{aligned}$

The oxidation state of O is -2.

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

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

Rearrange equation (5) for oxidation state of C

$(\text{oxidation state of C})=\left[-2(oxidation state of O)\right]$       ...... (6)

Substitute -2 for oxidation state of O in equation (6).

$\begin{aligned}\text{Oxidation state of C}&=[-2(-2)]\\&=+4\end{aligned}$

The oxidation state of C changes from -2 to +4. This shows it increases during the reaction and therefore $\textbf{C}_{2}\textbf{H}_{5}\textbf{OH}$ acts as a reducing agent.

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: redox reaction, C2H5OH, Cr, C, CO2, Cr2O72-, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously.