Home
Class 12
CHEMISTRY
The cell potential for the given cell at...

The cell potential for the given cell at 298K `pt|H_(2) (g, "1 bar")| H^(+) (aq)|| Cu^(2+) (aq)| Cu(s)` is 0.31V. The pH of the acidic solution is found to be 3, whereas the concentration of `Cu^(2+)` is `10^(-x)M`. The value of x is _____ (Given: `E_(Cu^(2+)//Cu)^(Ɵ) = 0.34V and (2.303RT)/(F)= 0.06V`)

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise CHEMISTRY |60 Videos

  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise PART : CHEMISTRY|30 Videos

  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise Chemistry Section A|20 Videos

  • JEE MAINS 2021

    JEE MAINS PREVIOUS YEAR|Exercise Chemistry (Section B )|10 Videos

  • JEE MAINS 2023 JAN ACTUAL PAPER

    JEE MAINS PREVIOUS YEAR|Exercise Question|360 Videos

Similar Questions

Explore conceptually related problems

The standard emf for the cell reaction, 2Cu^(+)(aq)to2Cu(s)+Cu^(2+)(aq) is 0.36V at 298K . The equilibrium constant of the reaction is

The standard cell potential for the cell Zn|Zn^(2+) (1 M)|| Cu^(2+) (1 M)| Cu Given E_(Cu^(2+)//Cu)^(@) = 0.34 V and E_(Zn^(2+)//Zn)^(@) = - 0.76 V is

Calculate DeltaG^(@) for the reaction : Cu^(2+)(aq) +Fe(s) hArr Fe^(2+)(aq) +Cu(s) . Given that E_(Cu^(2+)//Cu)^(@) = 0.34 V , E_(Fe^(+2)//Fe)^(@) =- 0.44 V

The standard e.m.f. for the cell reaction. 2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq) is +0.36 V at 298K. The equilibrium constant of the reaction is

Calculate the e.m.f of the cell Mg(s)//Mg^(2+)(0.1 M)||Cu^(2+)(1.0xx10^(-3) M)//Cu(s) Given E_(Cu^(2+)//Cu)^(@)=+0.34 V and E_(Mg^(2+)//Mg)^(@)=-2.37 V

For the disproportion of copper: 2 Cu^(+) to Cu^(+2) + Cu E^(0) is :- Given E^(0) for Cu^(+2)//Cu is 0.34 V & E^(0) for Cu^(+2)//Cu^(+) is 0.15 V:

E^(@) for the electrochemical cell Zn(s)|Zn^(2+) 1 M (Aq.)||Cu^(2+) 1 M (aq.)|Cu(s) is 1.10 V at 25^(@)C . The equilibrium constant for the cell reaction, Zn(s) +Cu^(2+) (aq.) hArr Zn^(2+) (aq.)+Cu(s) Will be :