Apart from the electrons which are produced at the anode and consumed at the cathode inside the Daniell cell, there are four other substances in this electrochemical reactions, i.e. Cu^{2+}, Cu, Zn^{2+} and Zn. Thus, the battery potential is given by,
\Delta \varphi = – \frac{1}{zF_F} \sum\limits_{A}{\nu _{\alpha A} }\mu _A .
= – \frac{1}{zF_F} (ν_{+Cu^{2+} μ_{Cu^{2+}}} + ν_{+Cu μ_{Cu} }+ ν_{−Zn^{2+} μ_{Zn^{2+}}} + ν_{−Zn μ_{Zn}}) .
where the + sign labels the reduction occurring at the cathode and the − sign labels the oxidation occurring at the anode. According to the redox equations at equilibrium (8.96) and (8.95), the stoichiometric coefficients ν_{+Cu^{2+}} =−1, ν_{+Cu} =1, ν_{−Zn^{2+}} =1, ν_{−Zn} =−1 , and the electrovalence is z = 2. Thus, the battery potential reduces to relation (8.108),
\bar{\mu }_{Cu} – \bar{\mu }_{Cu^{2+}} -2 \bar{\mu }_e = 0 . (8.96)
– \bar{\mu }_{Zn} + \bar{\mu }_{Zn^{2+}} +2 \bar{\mu }_e = 0 . (8.95).
\Delta \varphi ≡ \varphi ^{(+)}_e – \varphi ^{(-)}_e = \frac{1}{2 F_F} \biggl(\Bigl(\mu ^{(+)}_{Cu^{2+}} – \mu _{Cu}\Bigr)- \Bigl(\mu ^{(-)}_{Zn^{2+}} – \mu _{Zn}\Bigr) \biggr) . (8.108)
\Delta \varphi = \frac{1}{2 F_F}\biggl((\mu _{Cu^{2+}} – \mu _{Cu}) – (\mu _{Zn^{2+}} – \mu _{Zn}) \biggr) .
According to relation (8.18),
A_α = − \sum\limits_{A=1}^{r}{\mu _A \nu _{\alpha A}} . (8.18)
\Delta \varphi = – \frac{1}{z F_F} \sum\limits_{A}{ \nu _{\alpha A }\mu _A} = \frac{A_α}{z F_F} .
In view of relation (8.16),
\Delta _\alpha G ≡ \frac{\partial G}{\partial \xi _\alpha } = \sum\limits_{A=1}^{r}{\mu _A \nu _{\alpha A}} . (8.16)
\Delta \varphi = – \frac{1}{z F_F} \sum\limits_{A}{ \nu _{\alpha A }\mu _A} = -\frac{Δ_αG}{z F_F} .
According to relation (8.51),
μ_A = h_A − T s_A . (8.51)
\Delta \varphi = – \frac{1}{z F_F} \sum\limits_{A}{ \nu _{\alpha A }\mu _A} = – \sum\limits_{A}{ν_{αA} (h_A − T s_A)} = – \frac{Δ_αH − T Δ_αS}{z F_F} .