In the concentration cell
`Pt (H_(2)) |(HA (0.1M)),(Na A (1M))||(HA (1 M)),(Na A (1M))| (H_(2)) Pt`
`(pK_(a) " of " HA = 4)`
Cell potential will be :

Explanation

Hydrogen electrodes have been set up in two acidic buffer solution.
For the buffer, `Ha(0.1M),NaA(1M)`,
`[H^(+)]=K_(a)([HA])/([A^(-)])=10^(-4)xx(0.1)/(1)=10^(-5)M`
for the buffer, `HA(1M),NaA(1M)`,
`[H^(+)]=10^(-4)xx(1)/(1)=10^(-4)M`
for concentration cell, `E_(cell)=(0.0591)/(n)"log"(c_(2))/(c_(1))=(0.0591)/(n)"log"(10^(-4))/(10^(-5))=0.0591Vcong=0.06V`