
Identify the oxidizing agent and the reducing agent in the following redox reaction.
\[7C{N^ - } + 2O{H^ - } + 2Cu{(N{H_3})_4}^{2 + } \to 2Cu{(CN)_3}^{2 - } + 8N{H_3} + CN{O^ - } + {H_2}O\]
(A) Oxidising agent: $Cu{(N{H_3})_4}^{2 + }$ ; Reducing agent: $CN{O^ - }$
(B) Oxidising agent: $Cu{(N{H_3})_4}^{2 + }$ ; Reducing agent: $C{N^ - }$
(C) Oxidising agent: $CN{O^ - }$ ; Reducing agent: $C{N^ - }$
(D) None of these
Answer
133.5k+ views
Hint: The compound that undergoes oxidation in the reaction is called the reducing agent and the compound that undergoes reduction during the reaction is called an oxidizing agent. Calculate the oxidation state of all atoms before and after the reaction in order to identify the oxidising and reducing agent.
Complete step by step solution:
We know that in a redox reaction, oxidation and reduction reaction occurs simultaneously. That means some compounds are reduced and some are oxidized. We will check which reagents are oxidising agents and which are reducing ones.
-Oxidising agents are the compounds that undergo reduction during the reaction. Reducing agents are the compounds that undergo oxidation during the reaction.
-To find the oxidation number of an atom in the molecule, we will add the oxidation numbers of all the atoms which are there in the molecule and that sum will be equal to the overall charge of the molecule.
So, for $C{N^ - }$,
Overall charge on $C{N^ - }$ = Oxidation number of C + Oxidation number of N
-1 = Oxidation number of C + (-3)
Oxidation number of C = +2
Now, in $CN{O^ - }$,
Overall charge on $CN{O^ - }$ = Oxidation number of C + Oxidation number of N + Oxidation number of O
-1 = Oxidation number of C + (-3) + (-2)
Oxidation number of C = +4
-So, we can see that carbon is oxidized from +2 to +4 oxidation state. So, we can say that oxidation of $C{N^ - }$ has occurred here in the reaction. So, $C{N^ - }$ is the reducing agent.
In $Cu{(N{H_3})_4}^{2 + }$,
Overall charge on $Cu{(N{H_3})_4}^{2 + }$= Oxidation number of Cu + 4(Overall charge on $N{H_3}$)
+2 = Oxidation number of Cu + 4(0)
The oxidation number of Cu = +2
Now, in $Cu{(CN)_3}^{2 - }$,
Overall charge on $Cu{(CN)_3}^{2 - }$= Oxidation number of Cu + 3(Overall charge on $C{N^ - }$)
-2 = Oxidation number of Cu + 3(-1)
The oxidation number of Cu = +1
-So here we can see that reduction of copper occurs here from +2 to +1 state. Therefore there is a reduction of $Cu{(N{H_3})_4}^{2 + }$ during the reaction. So, $Cu{(N{H_3})_4}^{2 + }$ is called the oxidizing agent.
Thus, we can say that the oxidizing agent is $Cu{(N{H_3})_4}^{2 + }$ and the reducing agent is $C{N^ - }$.
Therefore, the correct answer is (B).
Note: Remember that the process of losing electrons is called oxidation and the process of gain of the electrons is called reduction. Also, note that $N{H_3}$ is a neutral molecule and it does not have any charge on it. It forms bonds with the metal by the lone pair of nitrogen atoms.
Complete step by step solution:
We know that in a redox reaction, oxidation and reduction reaction occurs simultaneously. That means some compounds are reduced and some are oxidized. We will check which reagents are oxidising agents and which are reducing ones.
-Oxidising agents are the compounds that undergo reduction during the reaction. Reducing agents are the compounds that undergo oxidation during the reaction.
-To find the oxidation number of an atom in the molecule, we will add the oxidation numbers of all the atoms which are there in the molecule and that sum will be equal to the overall charge of the molecule.
So, for $C{N^ - }$,
Overall charge on $C{N^ - }$ = Oxidation number of C + Oxidation number of N
-1 = Oxidation number of C + (-3)
Oxidation number of C = +2
Now, in $CN{O^ - }$,
Overall charge on $CN{O^ - }$ = Oxidation number of C + Oxidation number of N + Oxidation number of O
-1 = Oxidation number of C + (-3) + (-2)
Oxidation number of C = +4
-So, we can see that carbon is oxidized from +2 to +4 oxidation state. So, we can say that oxidation of $C{N^ - }$ has occurred here in the reaction. So, $C{N^ - }$ is the reducing agent.
In $Cu{(N{H_3})_4}^{2 + }$,
Overall charge on $Cu{(N{H_3})_4}^{2 + }$= Oxidation number of Cu + 4(Overall charge on $N{H_3}$)
+2 = Oxidation number of Cu + 4(0)
The oxidation number of Cu = +2
Now, in $Cu{(CN)_3}^{2 - }$,
Overall charge on $Cu{(CN)_3}^{2 - }$= Oxidation number of Cu + 3(Overall charge on $C{N^ - }$)
-2 = Oxidation number of Cu + 3(-1)
The oxidation number of Cu = +1
-So here we can see that reduction of copper occurs here from +2 to +1 state. Therefore there is a reduction of $Cu{(N{H_3})_4}^{2 + }$ during the reaction. So, $Cu{(N{H_3})_4}^{2 + }$ is called the oxidizing agent.
Thus, we can say that the oxidizing agent is $Cu{(N{H_3})_4}^{2 + }$ and the reducing agent is $C{N^ - }$.
Therefore, the correct answer is (B).
Note: Remember that the process of losing electrons is called oxidation and the process of gain of the electrons is called reduction. Also, note that $N{H_3}$ is a neutral molecule and it does not have any charge on it. It forms bonds with the metal by the lone pair of nitrogen atoms.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Thermodynamics Class 11 Notes: CBSE Chapter 5
