Home
Class 12
CHEMISTRY
For the following cell, Zn(s)|ZnSO(4)(...

For the following cell,
`Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)||Cu(s)`
When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG`
(in J `"mol"^(-1)`)
[F is Faraday constant, R is gas constant] T is temperaure, `E^(@)("cell")=1.1V`

A

2.303 RT +1.1 F

B

1.1 F

C

2.303 RT -2.2 F

D

`-2.2 F`

Text Solution

Verified by Experts

The correct Answer is:
C
(c ) `DeltaG=DeltaG^(@)+2.303" RT "log_(10)Q,Q=([Zn^(2+)])/([Cu^(2+)])`
`=-2F(1.1)+2.303" RT "log_(10)10`
`=2.303" RT"-2.2" F"`
Promotional Banner

Topper's Solved these Questions

  • ALCOHOLS AND PHENOLS

    DINESH PUBLICATION|Exercise COMPLETION QUESTION|2 Videos

  • ALCOHOLS AND PHENOLS

    DINESH PUBLICATION|Exercise NCERT In-Text Questions|9 Videos

  • ALDEHYDES AND KETONES

    DINESH PUBLICATION|Exercise Interger|5 Videos

Similar Questions

Explore conceptually related problems

For the following cell , Zn_(s)|ZnSO_(4)((aq)) CuSO_(4)((aq)) |Cu(s) When the concentration of Zn^(2+) is 10 times the concentration of Cu^(2+) , the expression for DeltaG ("in " J mol^(-1) is [F is Faraday constant , R is gas constant , T is temperature , E_(cell)^(@) = 1.1V ]

Zn(s)+CuSO_4(aq) to ZnSO_4(aq) + Cu(s) The above equation is an example of:

Write Nernst equation for the following cell reaction : Zn(s) | Zn^(2+)(aq)|| Cu^(2+)(aq) | Cu(s)

Calculate EMF of following cell at 298 K Zn|ZnSO_(4) (0.1 M) ||CuSO_(4) (1.0 M)|Cu (s) if E_(cell) = 2.0 V

Consider the following cell reaction Cu(s)+2Ag^(+)(aq) rarr Cu^(2+) (aq) +2Ag(s) E_("cell")^(@)=0.46 V By boubling the concentration of Cu^(2+) , E_("cell") is