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

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 :

A

`9.65 xx 10^(4)`

B

Antilog `(24.08)`

C

`37.3`

D

`10^(37.3)`

Text Solution

Verified by Experts

The correct Answer is:
D
Anode
`Zn - 2e rarr Zn^(2+)`
Cathode
`Cu^(2+) +2e rarr Cu`
Net cell reaction
`Zn +Cu^(2+) rarr Zn^(2+) +Cu`
Applying Nernst equation
`E_(cell) = E_(cell)^(@) - (0.059)/(2) log.([ZN^(2+)])/([Cu^(2+)])`
When the cell is completely discharged `E_(cell) = 0`
`0 = 1.1 - (0.059)/(2) log.([ZN^(2+)])/([Cu^(2+)])`
`log.([ZN^(2+)])/([Cu^(2+)]) = (2 xx 1.1)/(0.059) = 37.3`
or `([ZN^(2+)])/([Cu^(2+)]) = 10^(37.3)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Consider a Galvenic cell, Zn(s) |Zn^(2+) (0.1M) ||Cu^(2+) (0.1M)|Cu(s) by what factor, the electrolyte in anodic half cell should be diluted to increase the emf by 9 mili volt at 298K .

Zn|Zn^(2+)(c_(1)) || Zn^(2+)(c_(2))|Zn . For this cell DeltaG is negative if :

Estimate the cell potential of a Daniel cell having 1.0Zn^(++) and originally having 1.0M Cu^(++) after sufficient NH_(3) has been added to the cathode compartment to make NH_(3) concentration 2.0M . Given K_(f) for [Cu(NH_(4))_(4)]^(2+) = 1xx 10^(12), E^(@) for the reaction, Zn +Cu^(2+) rarr Zn^(2+) +Cu is 1.1V

Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe. The EMF of the above cell is 0.2905 . The equilibrium constant for the cell reaction is

50mL of 0.1M CuSO_(4) solution is electrolysed with a current of 0.965A for a period of 200sec. The reactions at electrodes are: Cathode: Cu^(2+) +2e^(-) rarr Cu(s) Anode: 2H_(2)O rarr O_(2) +4H^(+) +4e . Assuming no change in volume during electrolysis, calculate the molar concentration of Cu^(2+),H^(+) and SO_(4)^(2-) at the end of electrolysis.