Home
Class 12
CHEMISTRY
The magnitude ( but not the sign ) of th...

The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
If a half call `X|X^(2)(0.1M)` is connected to another half cell `Y|Y^(2+)(1.0M)` by means of a salt bridge and an external circuit at `25^(@)C`, the cell voltage would be

A

`0.06V`

B

`0.12V`

C

`0.62V`

D

`0.72V`

Text Solution

Verified by Experts

The correct Answer is:
c
When `X` is connected to `SHE`, electons flow from `X` to `SHE`. This means that `X` is acting as anode and `SHE` as cathode and its oxidation potential is positive. Also, the reduction potential of `Y` is greater than the reduction potential of `X(` as electrons flow from `X` to `Y).`
`implies Y^(2)+2e^(-) rarr Y(s)`
`E^(c-)._(Y^(2+)|Y)=0.34V,`
`X^(2+)+2e^(-) rarr X(s)`
`E^(c-)._(X^(2+)|X)=-0.25V`
Consider`: X(s)|X^(2+)(0.1M)||Y^(2+)(1.0M)|Y(s)`
`E^(c-)._(cell)=E^(c-)._(cell)-(0.059)/(2) log .([X^(2+)])/([Y^(2+)])`
`=[0.34-(-0.25)]-(0.059)/(2)log_(10).(0.1)/(1)=0.62V`
Promotional Banner

Similar Questions

Explore conceptually related problems

The reaction: Zn^(2+) + 2e^(-) rarr Zn has a standard reduction potential of -0.76 V . This means that

The standard electrode potentials E^(@) for the half cell reactions are as : ZntoZn^(2+)+2e^(-),E^(@)=0.76V FetoFe^(2+)+2e^(-),E^(@)=041V The EMF of the cell reaction Fe^(2+)+ZntoZn^(2+)+Fe is :

In the following half cell reactions, select the better reducing agent : Mg^(2+) + 2e^(-) rarr Mg, E^(@)= -2.37 V Fe^(3+) + 3e^(-) rarr Fe, E^(@) = -0.44 V

In a simple electrochemical cell, the half cell reduction reaction with their standard electrode potentials are : Pb(s)-2e^(-) rarr Pb^(2+)(aq) , E^(@)= - 0.30 V Ag(s) -e^(-) rarr Ag^(+) (aq), E^(@)= + 0.80 V Which of the following reaction takes place ?

If lim_(h rarr 0) (1 + (a)/(x) + (b)/(x^(2)))^(2x) = e^(2) , then the values of a and b, are :

The standard reduction potential of Fe^(3+), Fe^(2+)//Pt is + 0.771V . This half cell is connected with another half cell such that e.m.f. of the cell is 0.771V. What is the other half cell ?