Question

If `K_c` for the reaction
`Cu^(2+)(aq)+Sn^(2+)(aq)toSn^(4+)(aq)+Cu(s)`
at `25^(@)` C is represented as `2.6xx10^y` then find the value of y.
(Given:`E_(Cu^(2+)"|"Cu)^(@)=0.34V,E_(Sn^(4+)"|"Sn^(2+)^(@)=0.15V)`

Answer 1

The cell reaction :
`Sn^(2+)|Sn^(4+)||Cu^(2+)|Cu`
At the equilibrium point, `E_(cell)=0`
`E_(cell)=E Cu^(2+)|Cu-E_(Sn)^(4+)|Sn^(2+)=0`
`"or"[E^(@)Cu^(2+)|Cu+(0.0591)/(2)" log "[Cu^(2+)]]-[E^(@)Sn^(4+)|Sn^(2+)+(0.0591)/(2)"log"(Sn^(4+))/([Sn^(2+)])]=0`
`"or"[E_(Cu^(2+))^(@)|Cu-E_(Sn^(4+)|Sn^(2+))^(@)]+[(0.0591)/(2)"log"([Cu^(2+)][Sn^(2+)])/([Sn^(4+)])]=0`
`"or"(+0.34-0.15)-(0.0591)/(2)"log"([Sn^(4+)])/([Cu^(2+)][Sn^(4+)])=0`
`or 0.19-(0.0591)/(2)" log "K_(c )=0`
`log K_(c )=(0.19xx2)/(0.0591)=6.43`
`K_(c )="Anti "log (6.43)=2.69xx10^(6)(x=6)`