An electrochemical cell (galvanic cell) is designed by coupling Zn electrode with standard hydrogen electrode (SHE). Given, E Zn^++/Zn = 0.76\ V . i) Draw the galvanic cell. [2] ii) Write down…

i) Galvanic Cell Diagram

ii) Complete Cell Reaction

The anode (oxidation) half-reaction: $$\mathrm{Zn}(\mathrm{s})\to \mathrm{Zn}{}^{2+}(\mathrm{aq})+2\mathrm{e}{}^{-}$$ The cathode (reduction) half-reaction (SHE): $$2\mathrm{H}{}^{+}(\mathrm{aq})+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2(\mathrm{g})$$ Complete cell reaction (oxidation + reduction): $$\mathrm{Zn}(\mathrm{s})+2\mathrm{H}{}^{+}(\mathrm{aq})\to \mathrm{Zn}{}^{2+}(\mathrm{aq})+\mathrm{H}{}_2(\mathrm{g})$$

iii) Calculation of EMF

Given:

• $E_{{\text{Zn}}^{2+}/\text{Zn}}^{\circ }=-0.76\text{V}$
• $E_{{\text{H}}^{+}/{\text{H}}_2}^{\circ }=0.00\text{V}$ (standard hydrogen electrode)

The EMF (E°cell) of the galvanic cell is calculated as: $$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$$ $$E_{\text{cell}}^{\circ }=0.00\text{V}-(-0.76\text{V})=0.76\text{V}$$

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OR

Hess’s Law of Constant Heat Summation

Hess’s law states that the total enthalpy change ($\Delta H$) for a reaction is independent of the pathway and depends only on the initial and final states. Mathematically: $$\Delta H_{\text{reaction}}=\sum \Delta H_{\text{products}}-\sum \Delta H_{\text{reactants}}$$

Calculation of Heat of Formation of Ethanol ($\Delta H_f^{\circ }$ for $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}(\mathrm{l})$)

We use the following data:

1. $\mathrm{H}{}_2(\mathrm{g})+\frac{1}{2}\mathrm{O}{}_2(\mathrm{g})\to \mathrm{H}{}_2\mathrm{O}(\mathrm{l})$, $\Delta H=-286.2\text{kJ/mol}$
2. $\mathrm{C}(\mathrm{s})+\mathrm{O}{}_2(\mathrm{g})\to \mathrm{CO}{}_2(\mathrm{g})$, $\Delta H=-393.2\text{kJ/mol}$
3. $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}(\mathrm{l})+3\mathrm{O}{}_2(\mathrm{g})\to 2\mathrm{CO}{}_2(\mathrm{g})+3\mathrm{H}{}_2\mathrm{O}(\mathrm{l})$, $\Delta H=-1368\text{kJ/mol}$

Step 1: Write the formation reaction of ethanol: $$2\mathrm{C}(\mathrm{s})+3\mathrm{H}{}_2(\mathrm{g})+\frac{1}{2}\mathrm{O}{}_2(\mathrm{g})\to \mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}(\mathrm{l})$$

EthanolEthanol

Step 2: Use Hess’s law to derive $\Delta H_f^{\circ }$ for ethanol. Multiply the given reactions to match the stoichiometry of ethanol’s formation:

• For 2C(s) → 2CO₂(g): $2\times \mathrm{C}(\mathrm{s})+\mathrm{O}{}_2(\mathrm{g})\to \mathrm{CO}{}_2(\mathrm{g})$, $\Delta H=2\times (-393.2)=-786.4\text{kJ/mol}$
• For 3H₂O(l) → 3H₂(g) + 3/2O₂(g) (reverse of water formation): $3\times \mathrm{H}{}_2(\mathrm{g})+\frac{1}{2}\mathrm{O}{}_2(\mathrm{g})\to \mathrm{H}{}_2\mathrm{O}(\mathrm{l})$, $\Delta H=3\times 286.2=+858.6\text{kJ/mol}$
• For 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g) (reverse of ethanol combustion): $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}(\mathrm{l})+3\mathrm{O}{}_2(\mathrm{g})\to 2\mathrm{CO}{}_2(\mathrm{g})+3\mathrm{H}{}_2\mathrm{O}(\mathrm{l})$, $\Delta H=+1368\text{kJ/mol}$

Step 3: Combine the reactions to get the formation of ethanol: $$2C(s)+3H2(g)+\frac{1}{2}O2(g)->C2H5OH(l)$$ $$\Delta H_f^{\circ }=-786.4+858.6+1368=-261.8\text{kJ/mol}$$

Final Answer: The heat of formation of ethanol is $\Delta H_f^{\circ }=-261.8\text{kJ/mol}$.