Saved Bookmarks
| 1. |
Calculate the standard reduction potential of the following half cell S^(2-)|CuS|Cu Given: K_(sp)(CuS)=10^(-35)andE_(Cu^(2+)//Cu)^(@)=0.34V |
|
Answer» Solution :A cell can be assumed as follows: Anode: `Cu(s)toCu^(2+)+2e^(-),E^(@)=E_(1)` Cathode: `underline(CuS+2e^(-)toCu(s)+S^(2-),E^(@)=E_(2))` NET reaction: `CuStoCu^(2+)+S^(2-)` Applying Nernst equation to each of the above TWO electrode reaction `E_(1)=E_(Cu^(2+)//Cu)^(@)+(0.0591)/(2)LOG[Cu^(2+)]` `E_(2)=E_(S^(-)//CuS//Cu)^(@)-(0.0591)/(2)log[S^(2-)]` `E_(cell)=E_(2)-E_(1)` Wehn the cell reaction is in equilibrium, `E_(cell)=0` `thereforeE_(2)=E_(2)` i.e., `E_(S^(2-)//CuS//Cu)^(@)-(0.0591)/(2)log[S^(2-)]=E_(Cu^(2+)//Cu)^(@)+(0.0591)/(2)log[Cu^(2+)]` or `E_(S^(2-)//CuS//Cu)^(@)=E_(Cu^(2+)//Cu)^(@)+(0.0591)/(2)log[Cu^(2+)][S^(2-)]` `=E_(Cu^(2+)//Cu)^(@)+0.02955logK_(SP)(CuS)` `=0.34+0.02955log10^(-35)` `=0.34+0.02955(-35)=-0.694V` |
|