Sure! Let's solve the redox reactions for both cases A and B under acidic conditions using the provided half-reactions. The goal is to balance these reactions fully.
### Part A:
Reaction between [tex]\(\left( NH_4 \right)_2Fe\left( SO_4 \right)_2 \cdot 6 H_2O\)[/tex] and [tex]\(KMnO_4\)[/tex] under acidic conditions.
#### Step-by-Step Solution:
1. Identify the oxidation states:
- The relevant compound [tex]\((NH_4)_2Fe(SO_4)_2 \cdot 6 H_2O\)[/tex] contains iron in the +2 oxidation state (as [tex]\(Fe^{2+}\)[/tex]).
- Potassium permanganate [tex]\(KMnO_4\)[/tex] contains manganese in the +7 oxidation state (as [tex]\(MnO_4^-\)[/tex]).
2. Half-Reactions:
- Oxidation half-reaction:
Iron is oxidized from [tex]\(Fe^{2+}\)[/tex] to [tex]\(Fe^{3+}\)[/tex].
[tex]\[
Fe^{2+} \rightarrow Fe^{3+} + e^-
\][/tex]
- Reduction half-reaction:
Manganese is reduced from [tex]\(MnO_4^-\)[/tex] to [tex]\(Mn^{2+}\)[/tex] in acidic conditions.
[tex]\[
MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O
\][/tex]
3. Balance Electrons:
- Multiply the oxidation half-reaction by 5 to balance the electrons:
[tex]\[
5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-
\][/tex]
4. Combine the half-reactions:
[tex]\[
5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O
\][/tex]
Now, this is the fully balanced redox reaction for part A.
### Part B:
Reaction between [tex]\(K_2C_2O_4\)[/tex] and [tex]\(KMnO_4\)[/tex] under acidic conditions.
#### Step-by-Step Solution:
1. Identify the oxidation states:
- The compound [tex]\(K_2C_2O_4\)[/tex] contains oxalate [tex]\(C_2O_4^{2-}\)[/tex], in which carbon is in the +3 oxidation state.
- Potassium permanganate [tex]\(KMnO_4\)[/tex] contains manganese in the +7 oxidation state (as [tex]\(MnO_4^-\)[/tex]).
2. Half-Reactions:
- Oxidation half-reaction:
Oxalate is oxidized from [tex]\(C_2O_4^{2-}\)[/tex] to [tex]\(CO_2\)[/tex].
[tex]\[
C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-
\][/tex]
- Reduction half-reaction:
Manganese is reduced from [tex]\(MnO_4^-\)[/tex] to [tex]\(Mn^{2+}\)[/tex] in acidic conditions.
[tex]\[
MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O
\][/tex]
3. Balance Electrons:
- Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons:
[tex]\[
5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-
\][/tex]
[tex]\[
2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O
\][/tex]
4. Combine the half-reactions:
[tex]\[
5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O
\][/tex]
Now, this is the fully balanced redox reaction for part B.
These solutions provide the complete balanced reactions for the given compounds under acidic conditions.
