The standard reduction potential of `Cu^(2+)|Cu and Ag^+|Ag` electrodes are 0.337 and 0.799 volt respectively. Construct a galvanic cell using these electrodes so that its standard emf is positive. For what concentration of `Ag^(+)` will the emf of the cell at `25^@ C` be zero if the concentration of `Cu^(2+)` is 0.01M?

Options

A. `15.23 xx 10^(-9)M`
B. `1.523 xx 10^(-9)M`
C. `3.846 xx 10^(-9)M`
D. `38.46 xx 10^(-9)M`

Answer

B

Explanation

Given `E_(Cu^(2+)//Cu)^(@)= 0.337` volt and `E_(Ag^(+)//Ag)^(@)= 0.799` volt. The standard emf will be positive if `Cu//Cu^(2+)` is anode and `Ag^(+)//Ag` is cathode. The cell can be represented as: `Cu|Cu^(2+)||Ag^(+)|Ag`
The cell reaction is, `Cu+ 2Ag^(+) rarr Cu^(2+) + 2Ag`
`E_("cell")^(@)`= Oxide potential of anode + Red potential of cathode `= -0.337 + 0.799= 0.462` volt
Applying the Nernst equation, `E_("cell") = E_("cell")^(@) - (0.0591)/(2) "log" ([Cu^(2+)])/([Ag^(+)]^(2)0`
When, `E_("cell")= 0`
`E_("cell")^(@) =(0.0591)/(2) "log" ([Cu^(2+)])/([Ag^(+)]^(2)) or "log" ([Cu^(2+)])/([Ag^(+)]^(2))= (0.462 xx 2)/(0.0591) = 15.6345`
`([Cu^(2+)])/([Ag^(+)]^(2))= 4.3102 xx 10^(15),, [Ag^(+)]^(2)= (0.01)/(4.3102 xx 10^(15)`
`=0.2320 xx 10^(-17) = 2.320 xx 10^(-18) therefore [Ag^(+)] = 1.523 xx 10^(-9)M`