Estimate the `E^(c-)` reduction for `Cu|CuS` electrode.
Given `: K_(sp) ` of `CuS=8.0 xx 10^(-36),E^(c-)._((Cu|Cu^(2+)))=-0.34V`

Options

A. -0.71V
B. 0.71V
C. -1.42V
D. 1.42V

Answer

A

Explanation

Half cell reactions are
Anode reaction `:`
Cathode reaction `:
Cell reaction `:`
`E ^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)`
`=E^(c-)._((Cus|Cu))-E^(c-)._((Cu^(2+)|Cu))`
`=x-0.34V`
Cell is in equilibrium, therefore, `E_(cell)=0` .
`:. E_(cell)=E^(c-)._(cell)-(0.059)/(2)log[Cu^(2+)][S^(2-)]`
`0=(x-0.34 V)-(0.06)/(2)log K_(sp)(` Take`0.059~~0.06)`
`:. E^(c-)._(cell)=(0.06)/(2)log K_(sp)`
`-0.03log (8xx10^(-36))`
`=0.03[3log 2 -36]`
`=0.03[3xx0.3-36]`
`=0.03xx-35.1=-1.053V~~1.05V`
`:. (x-0.34)=-1.05V`
`x=-0.71V`
`E^(c-)._((CuS|Cu))(` standard reduction potential `)=-0.71V`