Galvanic cells generate electrical energy at te expense of a spontaneous redox reaction. In an electrode concentration cell two like electrodes having different concentrations, either because they are gas electrodes operating at different pressures or because they are amalgams (solutions in mercury) with different concentratiions are dipped into the same solution.
Eg. An example is a cell composed of two chlorine electrodes with different pressure of `Cl_(2)`:
`Pt_(L)|Cl_(2)(P_(L))|HCl(aq)|Cl_(2)(P_(R))|Pt_(R)`
Where `P_(L)` and `P_(R)` are the `Cl_(2)` pressure at the left and right electrodes.
Calculate the EMF of the electrode concentration cell represented by :
`Hg-Zn(c_(1))|Zn^(+2)(aq)|Hg-Zn(c_(2))`
A cell represented by the following cell diagram is generated to measure the solubility product of `AgCl`.
`Ag|Ag^(+)(aq)|Cl^(-)(aq)|AgCl(s)|Agg`
`K_(sp)^(0)(AgCl)` at `298K` and `1` bar pressure is:
Given:
`AgCl+e^(-)toAg+Cl^(-) E^(0)=0.22V`
`Ag^(+)+e^(-)toAg E^(0)=0.79V`
(Take `sqrt(10)=3.16` & `(RT.(2.303))/F=0.06`)

Options

A. `3.16xx10^(-10)`
B. `3.16xx10^(-11)`
C. `3.16xx10^(-9)`
D. data insufficient

Answer

A

Explanation

Oxidation halt -cell reaction
`Ag(s)toAg^(+)(aq)+e^(-)`
Reduction half-cell reaction
`AgCl(s)+e^(-)toAg(s)+Cl^(-)(aq)`
Net cell Reaction:
`AgCl(s)toAg^(+)(aq)+Cl^(-)(aq)`
At equilibrium:
`E_("cell")=0impliesE_("cell")^(0)=(RT)/(nF) ln K_(sp)`
and `E_("cell")^(@)=E_(Cl^(-)//AgCl//Ag)^(0)=E_(Ag^(+)//Ag)^(0)`
`=-0.57V`
`K_(sp)=10^(-10)xxsqrt(10)`
`=3.16xx10^(-10)`