To determine the voltage of an electrolytic cell with copper and magnesium electrodes, we need to use their standard reduction potentials. The relevant half-reactions and their potentials are:
1. Copper (Cu²⁺/Cu):
[tex]\( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \)[/tex]
(E° = +0.34 V)
2. Magnesium (Mg²⁺/Mg):
[tex]\( \text{Mg}^{2+} + 2e^{-} \rightarrow \text{Mg} \)[/tex]
(E° = -2.37 V)
In an electrolytic cell, the cathode is where the reduction occurs, and the anode is where the oxidation occurs. To find the overall cell potential (E_cell), we need to identify which electrode serves as the anode and which serves as the cathode. Typically, reduction occurs at the electrode with the higher (more positive) reduction potential.
Thus:
- Copper (Cu²⁺/Cu) will be reduced at the cathode,
- Magnesium (Mg²⁺/Mg) will be oxidized at the anode.
The cell potential for an electrolytic cell is calculated using the formula:
[tex]\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \][/tex]
Here:
- E_cathode (Copper) = +0.34 V
- E_anode (Magnesium) = -2.37 V
Substituting these values into our formula:
[tex]\[ E_{\text{cell}} = 0.34\, \text{V} - (-2.37\, \text{V}) \][/tex]
[tex]\[ E_{\text{cell}} = 0.34\, \text{V} + 2.37\, \text{V} \][/tex]
[tex]\[ E_{\text{cell}} = 2.71\, \text{V} \][/tex]
Therefore, the voltage of the electrolytic cell with copper and magnesium electrodes is 2.71 V. The correct answer is:
D. 2.71 V
