Home
Class 12
CHEMISTRY
In the cell Zn|Zn^(2+)||Cu^(2+)|Cu, the ...

In the cell `Zn|Zn^(2+)||Cu^(2+)|Cu`, the negaitve terminal is

A

Cu

B

`Cu^(2+)`

C

`Zn`

D

`Zn^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
C
In an electrochemical cell, anode (Zn) is a negative terminal
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NCERT FINGERTIPS ENGLISH|Exercise HOTS|10 Videos

  • ELECTROCHEMISTRY

    NCERT FINGERTIPS ENGLISH|Exercise EXEMPLAR PROBLEMS|17 Videos

  • COORDINATION COMPOUNDS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

Consider the cell Zn|Zn^(2+) || Cu^(2+)|Cu. If the concentration of Zn and Cu ions are doubled, the emf of the cell.

A variable, opposite external potential (E_(ext) is applied to the cell : Zn|Zn^(2+)(1M)||Cu^(2+) (1M)|Cu, of potential 1.1 V.respectively electrons flow from :

For the cell Zn(s)|Zn^(2+)||Cu^(2+)|Cu(s) , the standard cell voltage, E^(c-)._(cell) is 1.10V . When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.98V . One possible explanatino for the observed voltage is

A Daniell cell : Zn|Zn^(2+)||Cu^(2+)|Cu with E_(cell)=1.1V is given. Is this a spontaneous cell?

For the cell : Zn"|"Zn^(2+) (a=1)"||"Cu^(2+) (a=1)"|"Cu Given that E_(Zn//Zn^(2+))=0.761 V, E_(Cu^(2+)//Cu)=0.339V (i) Write the cell reaction. (ii) Calculate the emf and free energy change at 298 K involved in the cell. [Faraday's constant F = 96500 coulomb eq^(-1) ]

The emf of the cell Zn|Zn^(2+) (1 M)||Cu^(2+)|Cu(1M) is 1.1 volt. If the standard reduction potential of Zn^(2+)|Zn is -0.78 volt, what is the standard reduction potential of Cu^(2+)|Cu

Assertion : For the Daniel cell, Zn//Zn^(2+)||Cu^(2+)|Cu" with "E_(cell)=1.1V results into flow of electron from cathode to anode. Reason : Zn is deposited at anode and Cu is dissolved at cathode.

Consider the cell : Zn|Zn^(2+)(aq)(1.0M)||Cu^(2+)(aq)(1.0M)||Cu Thee standard reduction potentials are 0.350V for Cu^(2+)(aq)+2e^(-)rarrCu and -0.763V for Zn^(2+)(aq)+2e^(-) rarr Zn a. Write the cell reaction. b. Calculate the EMF of the cell. c. Is the reaction spontaneous or not ?

For the cell, Zn(s)abs(Zn^(2+))abs(Cu^(2+))Cu(s) , the standard cell voltage, E^(0)""_(cell) is 1.10 V. When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.98 One possible explanation for the observed voltage is :

Calculate the emf of the cell. Zn∣ Zn^(2+)(0.001M)∣∣Cu^(2+) (0.1M)∣Cu .The standard potential of Cu/ Cu^(2+) half-cell is +0.34 and Zn/ Zn^(2+) is -0.76 V.