Consider a cell composed of two cells: (i). `Cu(s)Cu^(2+)(aq)` and (ii). `Ag(s)|Ag^(+)(aq)` (b). The cell potential when `[Cu^(2+)]=2M` and `|Ag^(+)|=
Consider a cell composed of two cells:
(i). `Cu(s)Cu^(2+)(aq)` and (ii). `Ag(s)|Ag^(+)(aq)`
(b). The cell potential when `[Cu^(2+)]=2M` and `|Ag^(+)|=0.05M`
[Given: `C_(Cu^(2+)//Cu)^(@)=+0.344V,E_(Ag^(+)//Ag)^(@)=+0.80V]`
1 Answers
Given: `E_(Cu^(2+)//Cu)^(@)=+0.344V,E_(Ag^(+)//Ag)=+0.80V]`
Asked: (a). `E_(cell)^(0)=?`
(b). The cell potential whe `[Cu^(2+)]=2M` and `[Ag^(+)]=0.05M`
Formula used: `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
(a). `E_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(Cu^(2+)//Cu)^(@)=+0.80V-0.34V=0.46V`
(b). `Cu(s)toCu^(2+)(aq)+2e^(-)`
`underline(2Ag^(+)(aq)+2e^(-)to2Ag(s))`
`underline(Cu(s)+2Ag^(+)(aq)toCu^(2+)(aq)+2Ag(s))`
`E_(cell)=E_(cell)^(0)-(0.0591)/(2)log(([Cu^(2+)])/([Ag^(+)]))=0.46V-(0.0591)/(2)log((2)/((0.05)^(2))`
`=+0.46V-(0.0591)/(2)log((2xx100xx100)/(5xx5)=+0.46V-(0.0591)/(2)log800`
`=+0.46V-(0.0591)/(2)xx2.9031=+0.46-0.0857=0.3743V`.