Question

Calculate the e.f.m of the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
[given that `E_(Cr^(3+)//Cr)^(@)=-0.75,E_(Fe^(2+)//Fe)^(@)=-0.45V]`

Answer 1

Given `E_(Cr^(3+)//Cr)^(@)=-0.75,EK_(Fe^(2+)//Fe)^(@)=-0.45V`
Asked: e.m.f. of the cell: `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
Formula used:
(a). `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)`
(b). `E_(cell)-E_(cell)^(@)-(0.0591)/(n)log((["product"])/(["reactant"]))`
Explanation `E_(cell)^(@)="standard e.m.f. of cell"`
`E_(cell)=e.m.f.` of cell
Substitution and calculation
cell reaction `2Cr(s)+3Fe^(2+)(aq)to2Cr^(3+)(aq)+3Fe(s)`
`E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(2)=E_(Fe^(2+)//Fe)^(@)-E_(Cr^(3+)/Cr)^(@)=-0.45V-(-0.75)V=+0.30V`
`2Cr(s)to2Cr^(3+)(aq)+6e^(-)`
`underline(3Fe^(2+)(aq)+6e^(-)to3Fe(s))`
`underline(2Cr(s)+2Fe^(2+)(aq)to2C^(3+)(aq)+3Fe(s))`
`E_(cell)^(@)=E_(cell)^(@)-(0.0591)/(6)log(([Cr^(3+)])/([Fe^(2)]^(3)))`
`=0.30V-(0.0591)/(6)log((0.10)^(2))/((0.01)^(3)))=0.30V-(0.0591)/(6)xx4`
`=0.30V-(0.2364)/(6)=0.30-0.0394=0.2606`