Home
Class 12
CHEMISTRY
Consider the following cell reaction C...

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

A

doubled

B

halved

C

increases nut less than double

D

decreases by a small fraction

Text Solution

Verified by Experts

The correct Answer is:
D
`E_("cell")=E_("cell")^(@)-(RT)/(nF) "ln" ([Cu^(2+)])/([Ag^(+)]^(2))`
Doubling `[Cu^(2+)]`, decreases the emf by a small fraction.
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    BITSAT GUIDE|Exercise BITSAT Archives|13 Videos

  • D AND F BLOCK ELEMENTS

    BITSAT GUIDE|Exercise BITSAT Archives|10 Videos

  • GENERAL ORGANIC CHEMISTRY

    BITSAT GUIDE|Exercise BITSAT|13 Videos

Similar Questions

Explore conceptually related problems

In a cell reaction, Cu_((s))+ 2Ag_((aq))^(+) to Cu_((aq))^(2+) + 2Ag_((s)) E_"cell"^@ =+0.46 V . If the concentration of Cu^(2+) ions is doubled then E_"cell"^@ will be

Calculate the equilibrium constant of the reaction : Cu(s)+2Ag(aq) hArrCu^(2+)(aq) +2Ag(s) E^(c-)._(cell)=0.46V

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

What is the Gibbs energy of the following reaction? Zn(s)+Cu^(2+) (aq) rarr Zn^(2+)(aq)+Cu^(2+) (s), E_("cell")^(@)=1.1 V

The equilibrium constant (K) for the reaction Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s) , will be [Given, E_(cell)^(@)=0.46 V ]

Consider the following reaction: Cu(s)+2Ag^(+)(aq) to 2Ag(s)+Cu^(2+)(aq) Depict the galvanic cell in which the given reaction takes place.

The equilbrium constant for the reaction : Cu + 2 Ag^(+) (aq) rarr Cu(2+) (aq) +2 Ag, E^@ = 0. 46 V at 299 K is