Home
Class 12
CHEMISTRY
For the redox reaction: Zn(s) +Cu^(2+)...

For the redox reaction:
`Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)` taking place in a cell,
`E_(cell)^(@) is 1.10` volt. `E_(cell)` for the cell will be `(2.303 (RT)/(F) = 0l.0591)`

A

`1.07` volt

B

`0.82` volt

C

`2.14` volt

D

`1.80` volt

Text Solution

Verified by Experts

The correct Answer is:
A
Applying Nernst equation
`E_(cell) = E_(cell)^(@) =- (0.0591)/(2)log. |(Sn^(2+))/(Cu^(2+))|`
`=1.1 -(0.0591)/(2)log.(1)/(0.1)`
`1.1 - 0.02955`
`=1.07`
Promotional Banner

Similar Questions

Explore conceptually related problems

For the net cell reaction of the cell Zn(s) |Xn^(2+) ||Cd^(2+) |Cd(s) DeltaG^(@) in Kilojules at 25^(@)C is (E_(cell)^(@) = 0.360 V) :-

The measured e.m.f. at 25^(@)C for the cell reaction, Zn(s) +XCu^(2+)._((eq))(1.0M) rarr Cu(s) ._((aq))(0.1M) is 1.3 volt, calculate E^(@) for the cell reaction.

The standard emf for the cell cell reaction Zn + Cu^(2+) rarr Zn^(2+) + Cu is 1.10 volt at 25^@ C . The emf for the cell reaction when 0.1 M Cu^(2+) and 0.1 M ZN^(2+) solutions are used at 25^@ =C is .

Calculate the cell e.m.f. and DeltaG for the cell reaction at 298K for the cell. Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V at 298K . F = 96500 C mol^(-1) .

The emf of a cell corresponding to the reaction Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1 atm is 0.28 volt at 25^(@)C . Calculate the pH of the solution at the hydrogen electrode. E_(Zn^(2+)//Zn)^(@) =- 0.76 volt and E_(H^(+)//H_(2))^(@) = 0

Emf of a cell corresponding to the reaction:- Zn(s) +2H^(+) (aq) rarr Zn^(2+)(aq) [0.1M] +H_(2)(g) [1.0atm] is 0.49V at 25^(@)C, E_(Zn^(2+)//Zn)^(@) = - 0.76 V . The pH of solution in cathode chamber is.

In a cell that utilises the reaction. Zn(s) +2H^(+)(aq) rarr Zn^(2+) (aq) +H_(2)(s) addition of H_(2)SO_(4) to cathode compartement, will: (1) increase the E_(Cell) and shift equilibrium to the right (2) lower the E_(cell) and shift equilibrium to the right (3) lower the E_(cell) and shift equilibrium to the left (4) increase the E_(cell) and shift equilibrium to the left

Cu^(+) + e rarr Cu, E^(@) = X_(1) volt, Cu^(2+) + 2e rarr Cu, E^(@) = X_(2) X_(2) volt For Cu^(2+) + e rarr Cu^(+), E^(@) will be :

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