If `E_(Fe^(2+))^(@)//Fe = -0.441 V`
and `E_(Fe^(3+))^(@)//Fe^(2+) = 0.771 V`
The standard `EMF` of the reaction
`Fe+2Fe^(3+) rarr 3Fe^(2+)`
will be:

Options

A. 0.111 V
B. 0.330 V
C. 1.653 V
D. 1.212 V

Answer

D

Explanation

(d) Given that ` E_(Fe^(2+)//Fe)^(@)=-0.441V`
So, `Fe to Fe^(2+) +2e^(-),E^(@)=+0.441V`…(i)
and ` E_(Fe^(3+)//Fe^(2+))^(@)=-0.771V`
So, `2Fe^(3+)+2e^(-) to 2Fe^(2+), E^(@)=0.771V`…(ii)
Cell reaction
`{:( (i)" " Fe to Fe^(2+) +2e^(-)"," E^(@)=0.441 V),(underline((ii) 2Fe^(3+)+2e^(-) to Fe^(2+)"," E^(@)=+0.771 V" ")),(underline(" "Fe+2Fe^(3+) to 3Fe^(2+)"," E_(cell)^(@)=1.212 V" ")):}`
Alternative On the basis of cell rection following half-cell reactions are written
At anode
`Fe to Fe^(2+) +2e^(-) " ( oxidation)" `
At cathode
`2Fe^(3+)+2e^(-) to 2Fe^(2+) " (reduction)"`
So, `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`=(+0.771)-(-0.441)=+1.212 V`.