Home
Class 12
CHEMISTRY
Zn+Cu^(2+)(aq)toCu+Zn^(2+)(aq). Reacti...

`Zn+Cu^(2+)(aq)toCu+Zn^(2+)(aq)`.
Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])*E_(cell)^(@)=1.10V` ltb rgt `E_(cell)` will be 1.1591 V when :

A

`[Cu^(2+)//[Zn^(2+)]=0.01`

B

`[Zn^(2+)]//[Cu^(2+)]=0.01`

C

`[Zn^(2+)]//[Cu^(2+)]=0.1`

D

`[Zn^(2+)]//[Cu^(2+)]=1`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • GRAPHICAL INTERPRETATION

    GRB PUBLICATION|Exercise Multiple Objective Type|68 Videos
  • GRAPHICAL INTERPRETATION

    GRB PUBLICATION|Exercise Comprehension 1|1 Videos
  • GRAPHICAL INTERPRETATION

    GRB PUBLICATION|Exercise I.Electrochemistry|1 Videos
  • F-BLOCK ELEMENTS

    GRB PUBLICATION|Exercise Subjective type|7 Videos
  • HYDROCARBON (ALIPHATIC)

    GRB PUBLICATION|Exercise Previous years jee questions|28 Videos

Similar Questions

Explore conceptually related problems

Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq). Reaction quotient is Q=([Zn^(2+)])/([Cu^(2+)]) . Variation of E_(cell) with log Q is of the type with OA=1.10 V.E_(cell) will be 1.1591V when

For the redox reaction : Zn(s)+Cu^(2+)(aq) hArr Zn^(2+)(aq)+Cu(s) Reaction quotient (Q) =([Zn^(2+)(aq)])/([Cu^(2+)(aq)])=0.01 . What will be the value of E_(cell) ? Given that OA=1.10 V.

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

Consider the following reaction, Zn(s)+Cu^(2+) (0.1 M) rarr Zn^(2+) (1 M)+Cu(s) above reaction, taking place in a cell, E_("cell")^(@) is 1.10 V. E_("cell") for the cell will be (2.303 (RT)/(F)=0.0591)

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

Find out the E_("cell")^(@) from the given data (a) Zn|Zn^(+2)|| Cu^(+2)| Cu, E_("cell")^(@) = 1.10V (b) Cu| Cu^(+2)|| Ag^(+)|Ag, E_("cell")^(@) = 0.46V ( c) Zn|Zn^(+2)||Ag^(+) | Ag, E_("cell")^(@) = ? (Given E_(Cu^(+2)//Cu)^(@) = 0.34V )