To find the voltage of a galvanic cell made with magnesium (Mg) and gold (Au), we need to consider the standard reduction potentials for the two elements.
The standard reduction potential for magnesium (Mg²⁺/Mg) is [tex]\(-2.37 \, \text{V}\)[/tex] and for gold (Au³⁺/Au) is [tex]\(+1.50 \, \text{V}\)[/tex].
In a galvanic cell, the overall cell voltage (E°_cell) is determined by the difference between the cathode (reduction) potential and the anode (oxidation) potential. For this cell:
- The reduction (cathode) half-reaction is for gold: [tex]\( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \)[/tex] with a potential of [tex]\( +1.50 \, \text{V} \)[/tex].
- The oxidation (anode) half-reaction is for magnesium: [tex]\( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)[/tex] with a potential of [tex]\( -2.37 \, \text{V} \)[/tex].
The voltage of the cell is calculated by:
[tex]\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \][/tex]
Substituting the given potentials:
[tex]\[ E°_{\text{cell}} = +1.50 \, \text{V} - (-2.37 \, \text{V}) \][/tex]
[tex]\[ E°_{\text{cell}} = 1.50 \, \text{V} + 2.37 \, \text{V} \][/tex]
[tex]\[ E°_{\text{cell}} = 3.87 \, \text{V} \][/tex]
Therefore, the voltage of the galvanic cell made with magnesium (Mg) and gold (Au) is [tex]\( 3.87 \, \text{V} \)[/tex].
None of the given multiple-choice answers (A, B, C, D) match the computed voltage of 3.87 V. This seems to be an error in the provided answer choices.
