Answer :
To determine which element is reduced in the reaction [tex]\( \text{Cd(s)} + \text{PbBr}_2(\text{aq}) \rightarrow \text{Pb(s)} + \text{CdBr}_2(\text{aq}) \)[/tex], we need to follow the oxidation states of the elements involved and identify which element experiences a decrease in oxidation state. Here's the detailed, step-by-step solution:
1. Identify the initial and final oxidation states of the elements:
- For Cadmium (Cd):
- In the elemental form [tex]\( \text{Cd(s)} \)[/tex], the oxidation state of cadmium is 0.
- In [tex]\( \text{CdBr}_2(\text{aq}) \)[/tex], cadmium is in the form of [tex]\( \text{Cd}^{2+} \)[/tex], so its oxidation state is +2.
- For Lead (Pb):
- In [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex], lead is in the form of [tex]\( \text{Pb}^{2+} \)[/tex], so its oxidation state is +2.
- In the elemental form [tex]\( \text{Pb(s)} \)[/tex], the oxidation state of lead is 0.
2. Determine the changes in oxidation states:
- Cadmium (Cd):
- Changes from 0 in [tex]\( \text{Cd(s)} \)[/tex] to +2 in [tex]\( \text{CdBr}_2(\text{aq}) \)[/tex].
- This is an increase in oxidation state, indicating that cadmium is oxidized.
- Lead (Pb):
- Changes from +2 in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex] to 0 in [tex]\( \text{Pb(s)} \)[/tex].
- This is a decrease in oxidation state, indicating that lead is reduced.
3. Identify the element that is reduced:
- Reduction involves a decrease in oxidation state.
- Therefore, lead (Pb) goes from +2 in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex] to 0 in [tex]\( \text{Pb(s)} \)[/tex], indicating that lead (Pb) is the element that is reduced.
4. Select the correct answer:
- The answer is "Pb in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex]".
So, the element that is reduced in the reaction [tex]\( \text{Cd(s)} + \text{PbBr}_2(\text{aq}) \rightarrow \text{Pb(s)} + \text{CdBr}_2(\text{aq}) \)[/tex] is:
[tex]\[ \boxed{\text{Pb in PbBr}_2(\text{aq})} \][/tex]
1. Identify the initial and final oxidation states of the elements:
- For Cadmium (Cd):
- In the elemental form [tex]\( \text{Cd(s)} \)[/tex], the oxidation state of cadmium is 0.
- In [tex]\( \text{CdBr}_2(\text{aq}) \)[/tex], cadmium is in the form of [tex]\( \text{Cd}^{2+} \)[/tex], so its oxidation state is +2.
- For Lead (Pb):
- In [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex], lead is in the form of [tex]\( \text{Pb}^{2+} \)[/tex], so its oxidation state is +2.
- In the elemental form [tex]\( \text{Pb(s)} \)[/tex], the oxidation state of lead is 0.
2. Determine the changes in oxidation states:
- Cadmium (Cd):
- Changes from 0 in [tex]\( \text{Cd(s)} \)[/tex] to +2 in [tex]\( \text{CdBr}_2(\text{aq}) \)[/tex].
- This is an increase in oxidation state, indicating that cadmium is oxidized.
- Lead (Pb):
- Changes from +2 in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex] to 0 in [tex]\( \text{Pb(s)} \)[/tex].
- This is a decrease in oxidation state, indicating that lead is reduced.
3. Identify the element that is reduced:
- Reduction involves a decrease in oxidation state.
- Therefore, lead (Pb) goes from +2 in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex] to 0 in [tex]\( \text{Pb(s)} \)[/tex], indicating that lead (Pb) is the element that is reduced.
4. Select the correct answer:
- The answer is "Pb in [tex]\( \text{PbBr}_2(\text{aq}) \)[/tex]".
So, the element that is reduced in the reaction [tex]\( \text{Cd(s)} + \text{PbBr}_2(\text{aq}) \rightarrow \text{Pb(s)} + \text{CdBr}_2(\text{aq}) \)[/tex] is:
[tex]\[ \boxed{\text{Pb in PbBr}_2(\text{aq})} \][/tex]
