Home
Class 11
CHEMISTRY
A Galvanic cell consits of three compart...


A Galvanic cell consits of three compartment as shown in figure. The first compartment contain `ZnSO_4`(1M) and III compartment contain `CuSO_4`(1M). The mid compartment contain `NaNO_3` (1M). Each compartment contain 1L solution:
`E_(Zn^(2+)//Zn)^(@)=-0.76`, `E_(Cu^(2+)//Cu)^(@)=+0.34`,
The concertation of `Zn^(2+)` in first compartment after passage of 0.1 F charge will be:

A

1M

B

1.05M

C

1.025M

D

0.5M

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - One Or More Answers Are Correct|2 Videos
  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 2|3 Videos
  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

E^(@)Zn^(2+)//Zn = -0.77V . What is the 'E' value of the electrode containing 0.01M Zn^(2+) ions

5L of 0.1 M solution of sodium Carbonate contains

The emf of galvanic cell of the reaction Zn+Cu^(2+)rarr Zn^(2+) + Cu is (E^(@)Zn^(2+)//Zn " is" -0.76V " and " Cu^(2+)//Cu " is " +0.34V)

Define emf. Calculat the emf of the following galvanic cell : Zn_((s))+Cu_((aq))^(+2)to Zn_((aq))^(+2)+Cu_((s)) E_(zn^(+2//Zn))^(0) =0.76V(anode) ,E_(cu^(+3//Cu))^(0)=+0.34 (Cathode)

Calculate the concentration of silver ions in the cell constructed by using 0.1 M concentration of Cu^(2+) and Ag^(+) ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V. [E _(Ag^(2+)//Ag)=0.80V,E_(Cu^(2+)//Cu)=+0.34V]

What is nernst equation. Calculate the EMF of the galvanic cell construted fro these electrodes. Zn//Zn^(2+) (1.0M)"||"Cu^(2+)(1.0M)//Cu E^(@)Zn^(2+)//Zn= -0.766V, E^(@)Cu^(2+)//Cu= +0.327V

The emf (in V) of a Daniell cell containing 0.1 M ZnSO_(4) and 0.01 M CuSO_(4) solutions at their respective electrodes is (E_((Cu^(2+))/(Cu))^@ = +0.34 V, E_((Zn^(2+))/(Zn))^@ = -0.76 V)

Calculate the EMF of Zn//Zn^(2+)(0.1)"//"Cu^(2+)(0.1)//Cu E^(@) of Zn^(2+)//Zn = 0.762V , E^(@) of Cu^(2+)//Cu = +0.337V

Calculate the potential of a Zn -An^(2+) electrode in which the molarity of Zn ^(2+) is 0.001M. Given that E_(Zn^(2+)//Zn)^(0)=-0.76V R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).