Question

Calculate the EMF of the electrode concentration cell Hg-Zn `(c_(1)M)|(Zn^(2+)(c M)|` Hg-Zn `(c_(2)M)` at `25^(@)C`, if the concentration of the zinc amalgam are 2 g per 100 g of mercury and 1g per 100 g of mercury in anode and cathode half-cell respectively.
A. `6.8xx10^(-2)V`
B. `8.8xx10^(-3)V`
C. `5.7 xx 10^(-2)V`
D. `7.8xx10^(-3)V`

Answer 1

Correct Answer - B
The reactions at the two half-cells are
`{:("Ar cathode : "Zn^(2+)(c)+2e^(-)rarrZn(c_(2))),("At anode : "Zn(c_(1))rarrZn^(2+)(c)+2e^(-)),(bar("Net cell reactgion:"Zn(c_(1))rarr)Zn(c_(2))):}`
Applying Nernst equation to the net cell reaction gives
`E_(cell)=E_(cell)^(0)-(0.059)/(2)log.(c_(2))/(c_(1))=`
`(0.059)/(2)log.(c_(1))/(c_(2))("since E"_(cell)^(@)=0)`
`E_(cell)=(0.059)/(2)log.((2//65.4)/(1//65.4))=8.8xx10^(-3)v`