Question

Given `E_(Cr^(3+)//Cr)^(@)=-0.72V`, and
`E_(Fe^(2+)//Fe)^(@)=-0.42V`
The potential for the cell.
`Cr|Cr^(3+)(0.1M)||Fe^(2+)( 0. 01M)Fe` is
A. 0.072V
B. 0.3850V
C. 0.770V
D. 0.270V

Answer 1

Correct Answer - C
`E_("cell")^(@)=-0.42-(-0.72)=0.3V`
The cell reaction is
`2Cr+3Fe^(3+)r arr2Cr^(3+)+3Fe,n=6`
`E_("cell")=E_("cell")^(@)-(0.0591)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`=0.30-(0.591)/ (6) "log"((0.1)^(2))/((0.01)^(3))`
`=0.30-(0. 0 5 91)/(6)"log"10^(4)`
`=0.30-(0.0591xx 4)/(6)`
`=0.3 - 0.04=0.2 6V`.