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In the concentration cell, Pt(H(2)) |HA ...

In the concentration cell,` Pt(H_(2)) |HA (0.1M)|| HA(IM) NaA (IM)||NaA (IM)| (H_(2))Pt, (pK_(a) of HA = 4)` The cell potential will be:

A

0.03V

B

0.06V

C

`-0.06V`

D

`-0.03V`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell") = (0.0591)/(1) "log"_(10) ([H^(+)]"Cathode")/([H^(+)]"Anode"), E_("cell")= 0.06` [pH Anode- pH Cathode]…(i)
pH Anode `=pK_(a)+ log [HA] = 4+"log" (0.1)/(1)=3`, pH Cathode =4 (from eq 1) `E_("cell")= -0.06V`
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