Home
Class 12
CHEMISTRY
Given standard E^(c-): Fe^(3+)+3e^(-)r...

Given standard `E^(c-):`
`Fe^(3+)+3e^(-)rarr Fe, E^(c-)=-0.036`
Fe^(2+)+2e^(-)rarr Fe," "E^(c-)=-0.440V`
The `E^(c-)` of `Fe^(3+)+e^(-) rarr Fe^(2+) ` is

A

`-0.476V`

B

`-0.404V`

C

`0.404V`

D

`0.772V`

Text Solution

Verified by Experts

The correct Answer is:
d
`Fe^(3+)+3e^(-) rarr Fe(E_(1))" "(i)`
`Fe^(2+)+2e^(-) rarr Fe(E_(2))" "(ii)`
Net equation
`Fe^(3+)+e^(-) rarr Fe^(2+)(E_(3))" "(iii)`
`[` Obtained by `Eqs. (i)-(ii)]`
`E_(3)=(n_(1)E_(1)-N_(2)e_(2))/(n_(3))=(3(-0.036)-2(-0.0440))/(1)`
`=0.772V`
Promotional Banner

Similar Questions

Explore conceptually related problems

From the following data at 25^(@)C Cr^(3+)(aq.)+3e^(-)rarr Cr^(2+)(aq.) , E^(@)= -0.424V Cr^(2+)(aq.)+2e^(-) rarr Cr(s), " "E^(@)= -0.900V Find E^(@) at 25^(@)C for the reaction : Cr^(3+)+3e^(-) rarr Cr(s)

Given below are the half cell reactions : Mb^(2+)+2e^(-) rarr Mn, E^(@)= -1.18 V 2(Mn^(3+)+e^(-)rarr Mn^(2+)), E^(@)= +1.51V The E^(@) for 3Mn^(2+) rarr Mn + 2 Mn^(3+) will be

The rusting of iron takes place as follows : 2H^(+) + 2e^(-) + (1)/(2)O_(2)(g) rarr H_(2)O (l) E^(@) = + 1.23 V Fe^(2+) + 2e^(-) rarr Fe(s) E^(@) = - 0.44 V Calculate Delta G^(@) for the net process

Given E_(Fe^(3+)//Fe)^(@) = -0.36 V, E_(Fe^(2+)//Fe)^(@)= - 0.439V . The value of standard electrode potential for the change , Fe^(3+)+e^(-) rarr Fe^(2+) will be

The electrode potential of some half cell reactions are : Cl_(2) + 2e^(-) rarr 2 Cl^(-)(aq), E^(@) = + 1.36 V O_(2)+2H^(+)(aq) + 2e^(-) rarr H_(2)O_(2) (aq) , E^(@) = + 0.65 V F_(2)+2e^(-) rarr 2F^(-)(aq), E^(@) = + 2.87 V H_(2)O+ 2e^(-) rarr H_(2)(g) + 2OH^(-) (aq), E^(@) = - 0.83 V The strongest oxidising agent is :

The standard reduction potential at 298 K for the following half cell reactions are given below Zn^(2+)(aq)+2e^(-) to Zn(s) " " E^(@)=-0.762V Cr^(3+)(aq)+3e^(-)toCr(s) " " E^(@)=-0.740V 2H^(+)(aq)+2e^(-) toH_(2)(g) " " E^(@)=0.0V Fe^(3+)(aq)+e^(-) toFe^(2+)(aq) " " E^(@)=0.77V . Which is the strongest reducing agent ?