Question

The Gibbs energy change (in J) for the given reaction at $\left[{\mathrm{Cu}}^{2+}\right]$$=$ $\left[{\mathrm{Sn}}^{2+}\right]$$=$$1$ $\mathrm{M}$ and $298\mathrm{K}$ is:

$\mathrm{Cu}\left(\mathrm{S}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}.\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}.\right)+\mathrm{Sn}\left(\mathrm{s}\right)$

$({{\mathrm{E}}^0}_{{\mathrm{Sn}}^{2+}|\mathrm{Sn}}=-0.16\mathrm{V},{{\mathrm{E}}^0}_{{}_{{\mathrm{Cu}}^{2+}|\mathrm{Cu}}}=0.34\mathrm{V},\mathrm{Take}\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1})$

Answer 1

Gibbs energy:

• Gibbs free energy is referred to as a thermodynamic potential in thermodynamics.
• The maximum amount of reversible work that one thermodynamic system can complete at constant pressure and temperature is also determined using this potential. Joules are another SI unit used to measure Gibbs's free energy.
• Furthermore, the reduction in Gibbs energy that occurs during a thermodynamic system's reversible transition from its starting state to its end state is comparable to the work that the system and its surroundings perform. But here, the impact of pressure forces is not taken into account.

Solution:

$\mathrm{Cu}\left(\mathrm{S}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}.\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}.\right)+\mathrm{Sn}\left(\mathrm{s}\right)$

$({{\mathrm{E}}^0}_{{\mathrm{Sn}}^{2+}|\mathrm{Sn}}=-0.16\mathrm{V},{{\mathrm{E}}^0}_{{}_{{\mathrm{Cu}}^{2+}|\mathrm{Cu}}}=0.34\mathrm{V},\mathrm{Take}\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1})$

$$\begin{gathered} {{\mathrm{E}}^{°}}_{\mathrm{cell}}=-0.16-0.34 \\ =-0.50 \\ ∆{\mathrm{G}}^{°}=-\mathrm{nF}{{\mathrm{E}}^{°}}_{\mathrm{cell}} \\ =-2\times 96500\times \left(-0.5\right) \\ =+96500 \\ ∆\mathrm{G}={\mathrm{G}}^{°}+\mathrm{RT}\mathrm{nQ} \\ =96500+(25/3)\times 298\times 2.303\log \left(1\right) \\ ∆\mathrm{G}=96500\mathrm{Joules}. \end{gathered}$$