Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AC Be sure your answer has the correct number of significant digits.
2Zn (s) +N₂ (g)+4H2O (1)→2Zn2+ (aq) +N₂H₄(aq)+4OH (aq)
The Correct Answer and Explanation is :
To calculate the standard reaction free energy (( \Delta G^\circ )) for the reaction:
[
2\text{Zn (s)} + \text{N}_2 \text{(g)} + 4\text{H}_2\text{O (l)} \rightarrow 2\text{Zn}^{2+} \text{(aq)} + \text{N}_2\text{H}_4 \text{(aq)} + 4\text{OH}^- \text{(aq)}
]
We use the relationship:
[
\Delta G^\circ = -n F E^\circ_{\text{cell}}
]
Steps:
- Write Half-Reactions and Reduction Potentials:
- Oxidation: ( \text{Zn (s)} \rightarrow \text{Zn}^{2+} \text{(aq)} + 2e^- ), ( E^\circ_{\text{oxidation}} = -(-0.76 \, \text{V}) = +0.76 \, \text{V} ) (standard reduction potential for ( \text{Zn}^{2+}/\text{Zn} )).
- Reduction: ( \text{N}2 \text{(g)} + 4\text{H}_2\text{O (l)} + 4e^- \rightarrow \text{N}_2\text{H}_4 \text{(aq)} + 4\text{OH}^- \text{(aq)} ), ( E^\circ{\text{reduction}} = -1.16 \, \text{V} ).
- Calculate ( E^\circ_{\text{cell}} ):
[
E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} = -1.16 \, \text{V} + 0.76 \, \text{V} = -0.40 \, \text{V}.
] - Determine ( n ) (number of moles of electrons):
The oxidation of 2 moles of Zn produces 4 moles of electrons, so ( n = 4 ). - Calculate ( \Delta G^\circ ):
[
\Delta G^\circ = -n F E^\circ_{\text{cell}} = -(4)(96485 \, \text{C/mol})(-0.40 \, \text{V})
]
[
\Delta G^\circ = +154,376 \, \text{J/mol} = +154.38 \, \text{kJ/mol}.
]
Explanation:
This calculation determines the standard Gibbs free energy change for the reaction under standard conditions. The negative sign in the equation ( \Delta G^\circ = -n F E^\circ_{\text{cell}} ) reflects the relationship between spontaneity and cell potential. Since the cell potential (( E^\circ_{\text{cell}} )) is negative, the reaction is non-spontaneous under standard conditions (( \Delta G^\circ > 0 )).
Proper use of significant figures ensures that the final result reflects the precision of the given data. Both ( E^\circ ) values are given to 2 decimal places, so the result for ( \Delta G^\circ ) is reported with 5 significant figures, consistent with the constants used.