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194 6 CHEMICAL EQUILIBRIUM
�e Nernst equation from part (i) is rearranged for E−○cell and the expres-
sion for γ± is substituted in, noting from inside the front cover that ln x =
ln 10 × log x.
E−○cell = Ecell +
RT
F
ln(γ2±b2
b−○ 2
) = Ecell +
2RT
F
[ln( b
b−○
) + ln 10 × log γ±]
= Ecell +
2RT
F
⎡⎢⎢⎢⎣
ln( b
b−○
) − ln 10 × A( b
b−○
)
1/2⎤⎥⎥⎥⎦
= (+0.4658 V) + 2 × (8.3145 JK−1mol−1) × (298.15 K)
96485Cmol−1
×
⎡⎢⎢⎢⎢⎣
ln(0.01 mol kg
−1
1 mol kg−1
) − ln 10 × (0.509) × (0.01 mol kg
−1
1 mol kg−1
)
1/2⎤⎥⎥⎥⎥⎦
= +0.223 V
Finally, note that E−○cell = E−○(R)−E−○(L) = E−○(AgCl/Ag, Cl−)−E−○(H+/H2).
Because E−○(H+/H2) = 0 (by de�nition), it follows that E−○(AgCl/Ag, Cl) =
E−○cell = +0.223 V
E6C.4(b) �e reduction half-reactions for the cell in question are
R: 2NO−3 (aq) + 4H+(aq) + 2e− → 2NO2(g) + 2H2O(l)
L: Zn2+(aq) + 2e− → Zn(s)
which reveal that ν = 2 for the given cell reaction. As explained in Section 6C.3(a)
onpage 219 themaximumnon-expansion (electrical)work that a reactionwhen
it advances by an in�nitesimal amount dξ at some composition is given by
dwe = ∆rG dξ. �e reaction Gibbs energy ∆rG therefore represents the work
done per mole of reaction, that is, when the reaction advances by ∆ξ = 1mol at
constant composition.�e reaction Gibbs energy is related to the cell potential
according to [6C.2–217], ∆rG = −νFEcell, so assuming that the cell is operating
under standard conditions the electrical work that can be done (per mole of
reaction) is
dwe/dξ = ∆rG−○ = −2FE−○cell = −2 × (96485Cmol−1) × (−0.040 V)
= +7.7 kJmol−1
�e positive value indicates that work has been done on the system by the
surroundings.
E6C.5(b) �e Nernst equation [6C.4–221] is Ecell = E−○cell − (RT/νF) lnQ. If Q changes
from Q1 to Q2 then the change in cell potential is given by
Ecell,1 − Ecell,2 = [E−○cell −
RT
νF
lnQ2] − [E−○cell −
RT
νF
lnQ1] = −
RT
νF
ln(Q2
Q1
)