In a cell with [tex]\( \text{Zn} / \text{Zn}^{2+} \)[/tex] and [tex]\( \text{Cu} / \text{Cu}^{2+} \)[/tex] electrodes, we need to determine the redox reactions that take place.
To understand the process, let's consider the standard reduction potentials of zinc and copper:
- The standard reduction potential for [tex]\( \text{Zn}^{2+} / \text{Zn} \)[/tex] is [tex]\( -0.76 \, \text{V} \)[/tex].
- The standard reduction potential for [tex]\( \text{Cu}^{2+} / \text{Cu} \)[/tex] is [tex]\( +0.34 \, \text{V} \)[/tex].
In an electrochemical cell, the species with the higher reduction potential will undergo reduction, and the species with the lower reduction potential will undergo oxidation.
1. The [tex]\( \text{Cu}^{2+} / \text{Cu} \)[/tex] couple has a higher reduction potential (+0.34 V) compared to the [tex]\( \text{Zn}^{2+} / \text{Zn} \)[/tex] couple (-0.76 V). This means copper ions ([tex]\( \text{Cu}^{2+} \)[/tex]) are more likely to gain electrons and be reduced, while zinc metal (Zn) is more likely to lose electrons and be oxidized.
2. Therefore, in this cell, the chemical reactions occurring are:
- Oxidation at the anode: Zinc metal (Zn) loses two electrons to form zinc ions ([tex]\( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)[/tex]).
- Reduction at the cathode: Copper ions ([tex]\( \text{Cu}^{2+} \)[/tex]) gain two electrons to form copper metal ([tex]\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)[/tex]).
Given these reactions, we can conclude:
- Zinc (Zn) is oxidized, meaning Zn loses electrons.
- Copper ions ([tex]\( \text{Cu}^{2+} \)[/tex]) are reduced, meaning [tex]\( \text{Cu}^{2+} \)[/tex] gains electrons.
Therefore, the correct answer is:
d. [tex]\( \text{Cu}^{2+} \)[/tex] is reduced and Zn is oxidized.
