Home
Class 12
CHEMISTRY
The cell, Zn//Zn^(2+)(1M)"//"Cu^(2+)(1M)...

The cell, `Zn//Zn^(2+)(1M)"//"Cu^(2+)(1M)//Cu(E_(cell)^(@)=1.10V)` , was allowed to be completely concentration of `Zn^(2+)` to `Cu^(2+)` is

A

37.3

B

1037.3

C

`9.65 xx 10^(4)`

D

antilog (24.04)

Text Solution

Verified by Experts

The correct Answer is:
B
Ecell = `E^(@)cell -0.059/2 log ([Zn^(2+)]/[Cu^(2+)])`
Promotional Banner

Similar Questions

Explore conceptually related problems

The cell, Zn | Zn^(2+) (1M) || (1M) | Cu (E_(cell)^(@) = 1.10V) , was was allowed to be completely discharged at 298 K. The concentration of Zn^(2+) rightarrow CU^(2+)([([Zn^(2+)]/([Cu^(2+)])]) is :

The cell Zn//Zn^(2+) (1 M) ||Cu^(2+)(1M)//Cu(E_("cell")^(0) = 1.10 V) was allowed to be completely discharged at 298 K. The relative concentration of Zn^(2+) to Cu^(2+) (([Zn^(2+)])/([Cu^(2+)])) is 10^(x) . The value of x is : (take (2.303 RT)/(F) = 0.059 round off your answer up to one decimal)

The cell , Zn | Zn^(2+) (1M) || Cu^(2+) (1M) Cu (E_("cell")^@ = 1. 10 V) , Was allowed to be completely discharfed at 298 K . The relative concentration of 2+ to Cu^(2+) [(Zn^(2=))/(Cu^(2+))] is :

In the cell, Zn|Zn^(2+)"||"Cu^(2+)|Cu ________.

For the cell given, below Zn|Zn^(2)||Cu^(2+)|Cu,(E_(cell)-E_(cell)^(@)) is -0.12 V. it will be when