Home
Class 11
CHEMISTRY
When a solution of AgNO3 (1 M) is electr...

When a solution of `AgNO_3` (1 M) is electrolysed using Pt anode and Cu cathode, the products at two electrodes are?`E_(cu^(2+)|cu)^(@)=+0.34 "V" , E_(O_2,H^+|H_2O)^(@)=+1.23"V",E_(ag^+|Ag)^(@)=+0.8 "V"`

A

`CutoCu^(2+)` at anode, `Ag^+to Ag` at cathode

B

`H_2OtoO_2` at anode, `Cu^(2+)to Cu` at cathode

C

`H_2OtoO_2` at anode, `Ag^+to Ag` at cathode

D

`NO_3^-to NO_2` at anode, `Ag^+to Ag` at cathode

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of identifying the products formed at the anode and cathode during the electrolysis of a 1 M solution of AgNO3 using a platinum anode and a copper cathode, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Electrolyte Components**: - When AgNO3 is dissolved in water, it dissociates into Ag⁺ ions and NO3⁻ ions. Additionally, water dissociates into H⁺ and OH⁻ ions. - Therefore, the ions present in the solution are: Ag⁺, NO3⁻, H⁺, and OH⁻. 2. **Determine the Reactions at the Cathode**: - At the cathode, reduction occurs. The possible reduction reactions are: - Ag⁺ + e⁻ → Ag (E° = +0.80 V) - 2H⁺ + 2e⁻ → H2 (E° = 0.00 V) - Since the standard reduction potential for Ag⁺ to Ag is higher (+0.80 V) than that for H⁺ to H2 (0.00 V), Ag⁺ will be reduced to Ag. - **Product at Cathode**: Silver (Ag) is deposited. 3. **Determine the Reactions at the Anode**: - At the anode, oxidation occurs. The possible oxidation reactions are: - NO3⁻ → (oxidation, but nitrogen is already in its highest oxidation state of +5, so it cannot be oxidized further) - 4OH⁻ → 2H2O + O2 + 4e⁻ (E° = +0.40 V for OH⁻ oxidation) - Since the nitrate ions cannot be oxidized further, the hydroxide ions (OH⁻) will be oxidized to produce oxygen gas (O2). - **Product at Anode**: Oxygen (O2) is produced. 4. **Summarize the Products**: - At the cathode: Silver (Ag) is formed. - At the anode: Oxygen (O2) is produced. ### Final Answer: - **Products**: At the cathode, silver (Ag) is formed, and at the anode, oxygen (O2) is produced.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI ENGLISH|Exercise LEVEL-2|28 Videos

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI ENGLISH|Exercise LEVEL-3|37 Videos

  • DILUTE SOLUTION

    NARENDRA AWASTHI ENGLISH|Exercise leval-03|23 Videos

  • GASEOUS STATE

    NARENDRA AWASTHI ENGLISH|Exercise Subjective problems|15 Videos

Similar Questions

Explore conceptually related problems

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

Which of the following statements is correct? If E_(Cu^(2+)|Cu)^(@)=0.34V, E_(Sn^(2+)|Sn)^(@)=-0.136V,E_(H^+|H_2)^(@)=-0.0V

E_(Cu^(2+)//Cu)^(@) =0.43V E_(Cu^(+)//Cu)^(@)=0.55V E_(Cu^(2+)//Cu^(+))^(@) =

Can a solution of 1 M ZnSO_(4) be stored in a vessel made of copper ? Given that E_(Zn^(+2)//Zn)^(@) =-0.76V and E_(Cu^(+2)//Cu)^(@)=0.34 V

An aqueous solution containing 1M each of Au^(3+),Cu^(2+),Ag^(+),Li^(+) is being electrolysed by using inert electrodes. The value of standard potentials are : E_(Ag^(+)//Ag)^(@)=0.80 V,E_(Cu^(+)//Cu)^(@)=0.34V and E_(Au^(+3)//Au)^(@)=1.50,E_(Li^(+)//Li)^(@)=-3.03V will increasing voltage, the sequence of deposition of metals on the cathode will be :

Four different solution containing 1M each of Au^(+3), Cu^(+2), Ag^(+), Li^(+) are being electrolysed by using inert electrodes. In how many samples, metal ions would be deposited at cathode? ["Given :" E_(Ag^+//Ag)^(0) = 0.8, E_(Au^(+3)//Au)^(0) = 1.00 V E_(Cu^(+2)//Cu)^(0) = 0.34 V, E_(Li^(+)//Li)^(0)= -3.03 V ]

What must be concentration of Ag^(+) in an aqueous solution containing Cu^(2+) =1.0 M so that bot the metals can be deposited on the cathode simultaneously. Given that E_(Cu//Cu^(2+))^(0)=0.34V and E_(Ag^(+)//Ag)^(0)=0.812V,T=298K

Which is/are correct among the following? Given the half cell EMFs E_(Cu^(2+)//Cu)^(@)=0.337V, E_(Cu^(+)|Cu)^(@)=0.521V

Copper from copper sulphate solution can be displacesd by. (The standard reduction potentials of some electrodes are given below): E^(o)(Fe^(2+)//Fe) = - 0.44 V, E^(o) (Zn^(2+)//Zn) = - 0.76 V E^(o) (Cu^(2+)//Cu) = + 0.34 V: E^(o)(Cr^(3+)//Cr) = - 0.74 V E^(o) (H^(+)// 1/2H_(2)) = 0.00V

Home
Profile