The two half cells would have different value of H+ ion

..........HA1....<.................>H+.+A1-
HA2......<.........................>H+.+A2-
H+(for HA1)=(Ka1*C)1/2..M
H+(for HA2)=(Ka2*C)1/2..M

H+...+e-.....>1/2H2
writing for the half cell, EH+/H2=E0H+/H2-(0.0591/1)*log(pH2/[H+])
writing these two half cell reaction for both the solution and taking the value of standard reduction potential of H=0
E1H+/H2=0-(0.0591/1)*log(pH2/[H+]1)..[H+]=(Ka1*C)1/2…................(i)

E2H+/H2=0-(0.0591/1)*log(pH2/[H+]2)..where..[H+]=(Ka2*C)½..........(ii)
As pka1=3.so Ka=10-3

and pka2=5.so ka2=10-5

and so [H+] in HA1>[H+] in HA2 but the Ph2 =1atm in both the cases.
So to make the overall reaction feasible Ecell>0 and so to get this positive value, we should make solution of HA1 as cathode and so Ecell can be given as Ecell=Ecathode-Eanode=E1-E2>0
writing the cell notation:
H2(1atm)/H+(in HA2)//H+(in HA1)/H2(1atm)
So subtracting (ii) from (i),
we have Ecell=E1H+/H2-E2H+/H2=-(0.0591/1)*log(pH2/[H+]1)..+...(0.0591/1)*log(pH2/[H+]2)
=0.0591[log{[H+]1/[H+]2]=0.0591*log{Ka1C/Ka2C}1/2=0.0591*(1/2)*logKa1/Ka2
=0.0591V