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208 6 CHEMICAL EQUILIBRIUM
�e expression for Ka is rearranged to give aH+ = KaaBH+/aB and this is sub-
stituted into the Nernst equation to give
Ecell = E−○cell −
RT
F
ln (aCl−aH+) = E−○cell −
RT
F
ln( aCl
−aBH+Ka
aB
)
Replacing activities by aJ = γJ(bJ/b−○) [5F.14–185] gives
Ecell = E−○cell −
RT
F
ln((γCl−bCl−/b−○)(γBH+bBH+/b−○)Ka
(γBbB/b−○)
)
In this case bCl− = bBH+ = bB so the Nernst equation simpli�es to
Ecell = E−○cell −
RT
F
ln(γCl−γBH+
γB
× bKa
b−○
) = E−○cell −
RT
F
ln(γ2±bKa
b−○
)
where the mean activity coe�cient of the BH+ and Cl– ions is given by [5F.22–
187], γ± = (γCl−γBH+)1/2 and the neutral base B is assumed to be an ideal solute
so that γB = 1. Noting from inside the front cover that ln x = ln 10 log x, the
Nernst equation becomes
Ecell = E−○cell −
RT ln 10
F
log(γ2±bKa
b−○
)
= E−○cell −
RT ln 10
F
(2 log γ± + log(
b
b−○
) − pKa)
where pKa = − logKa has been used. Next the Davies equation [5F.30b–189],
log γ± = −A∣z+z−∣I1/2/(1+BI1/2)+CI, is used to substitute for log γ±.�e ionic
strength I is given by [5F.28–188], I = 1
2 ∑i z
2
i (b i/b−○), where z i is the charge
on ion species i and the sum extends over all the ions present in the solution.
In this case, bBH+ = bCl− = b, and bH+ is neglected because it will be much
smaller on account of the equilibrium involving the base B.�erefore the ionic
strength is
I = 1
2 (z
2
BH+b + z2Cl−b) /b−○ = 1
2 (1
2 × b + (−1)2 × b) /b−○ = b/b−○
and therefore
log γ± = −
A∣zBH+ × zCl− ∣I1/2
1 + BI1/2
+ CI = − A(b/b−○)1/2
1 + B(b/b−○)1/2
+ C ( b
b−○
)
Substitution of this expression into the Nernst equation derived above gives
Ecell = E−○cell −
RT ln 10
F
(2 [− A(b/b−○)1/2
1 + B(b/b−○)1/2
+ C ( b
b−○
)] + log( b
b−○
) − pKa)
which rearranges to
F(Ecell − E−○cell)
RT ln 10
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
y
= 2A(b/b−○)1/2
1 + B(b/b−○)1/2
− 2C ( b
b−○
) − log( b
b−○
) + pKa