Answer :
To determine which substance should be used as the anode in the redox reaction given, we need to follow the steps to understand which element is getting oxidized and which is getting reduced.
The electrochemical reaction is:
[tex]\[ Fe^{3+} + V^{2+} \rightarrow Fe^{2+} + V^{3+} \][/tex]
1. Identify the oxidation states of the elements involved:
- Iron (Fe) changes from [tex]\( Fe^{3+} \)[/tex] to [tex]\( Fe^{2+} \)[/tex].
- Vanadium (V) changes from [tex]\( V^{2+} \)[/tex] to [tex]\( V^{3+} \)[/tex].
2. Determine which species is reduced and which is oxidized:
- Reduction means a gain of electrons. In the reaction, [tex]\( Fe^{3+} \)[/tex] gains one electron to become [tex]\( Fe^{2+} \)[/tex]:
[tex]\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \][/tex]
- Oxidation means a loss of electrons. In the reaction, [tex]\( V^{2+} \)[/tex] loses one electron to become [tex]\( V^{3+} \)[/tex]:
[tex]\[ V^{2+} \rightarrow V^{3+} + e^- \][/tex]
3. Identify the anode and the cathode:
- Oxidation occurs at the anode. So, the substance that gets oxidized will be at the anode.
- Reduction occurs at the cathode.
4. Determine the substance used as the anode:
- Since vanadium (V) is oxidized (losing electrons) from [tex]\( V^{2+} \)[/tex] to [tex]\( V^{3+} \)[/tex], vanadium should be at the anode.
Hence, the correct substance to be used as the anode is:
[tex]\[ \text{vanadium because it loses electrons} \][/tex]
The electrochemical reaction is:
[tex]\[ Fe^{3+} + V^{2+} \rightarrow Fe^{2+} + V^{3+} \][/tex]
1. Identify the oxidation states of the elements involved:
- Iron (Fe) changes from [tex]\( Fe^{3+} \)[/tex] to [tex]\( Fe^{2+} \)[/tex].
- Vanadium (V) changes from [tex]\( V^{2+} \)[/tex] to [tex]\( V^{3+} \)[/tex].
2. Determine which species is reduced and which is oxidized:
- Reduction means a gain of electrons. In the reaction, [tex]\( Fe^{3+} \)[/tex] gains one electron to become [tex]\( Fe^{2+} \)[/tex]:
[tex]\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \][/tex]
- Oxidation means a loss of electrons. In the reaction, [tex]\( V^{2+} \)[/tex] loses one electron to become [tex]\( V^{3+} \)[/tex]:
[tex]\[ V^{2+} \rightarrow V^{3+} + e^- \][/tex]
3. Identify the anode and the cathode:
- Oxidation occurs at the anode. So, the substance that gets oxidized will be at the anode.
- Reduction occurs at the cathode.
4. Determine the substance used as the anode:
- Since vanadium (V) is oxidized (losing electrons) from [tex]\( V^{2+} \)[/tex] to [tex]\( V^{3+} \)[/tex], vanadium should be at the anode.
Hence, the correct substance to be used as the anode is:
[tex]\[ \text{vanadium because it loses electrons} \][/tex]
