Question

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 :
A. 0.03V
B. 0.06V
C. `-0.06V`
D. `-0.03V`

Answer 1

Correct Answer - C
`E_("cell") = (0.091)/(1) " log"_(10) ([H^(+)]" Cathode")/([H^(+)] " Anode")`
`E_("cell") = 0.06` [pH Anode `- pH` Cathode]....(1)
pH Anode `= pK_(a) +` log `[bar(HA)] = 4 + " log" (0.1)/(1) = 3`
pH Cathode = 4 (from eq.1)
`E_("cell") = - 0.06V`