Answer :
Let's tackle both parts of this problem step-by-step.
### (i) Balancing the Redox Equation Using Ion-Electron Method
First, let’s balance the given reaction using the ion-electron method. The reaction in question is:
[tex]\[ Zn + NO_3^- \rightarrow Zn^{2+} + NH_4^+ \][/tex]
Step 1: Write the oxidation and reduction half-reactions.
Oxidation half-reaction:
[tex]\[ Zn \rightarrow Zn^{2+} + 2e^- \][/tex]
Reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 2: Balance the atoms other than oxygen and hydrogen.
The zinc atoms in the oxidation half-reaction are already balanced.
Step 3: Balance oxygen atoms by adding [tex]\(H_2O\)[/tex] molecules.
In the reduction half-reaction:
There are no changes needed since it is already balanced for oxygen atoms.
Step 4: Balance hydrogen atoms by adding [tex]\(H^+\)[/tex] ions.
Again, the reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
The reduction half-reaction is balanced for hydrogen.
Step 5: Balance the charges by adding electrons.
The oxidation half-reaction:
[tex]\[ Zn \rightarrow Zn^{2+} + 2e^- \][/tex]
The reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 6: Ensure the number of electrons lost in oxidation equals the number of electrons gained in reduction.
To balance the electrons, we need to multiply the oxidation half-reaction by 4 and the reduction half-reaction by 1:
[tex]\[ 4(Zn \rightarrow Zn^{2+} + 2e^-) \][/tex]
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Each side of the reaction should now have 8 electrons:
[tex]\[ 4Zn \rightarrow 4Zn^{2+} + 8e^- \][/tex]
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 7: Combine the half-reactions and simplify.
Combining these, we get:
[tex]\[ 4Zn + NO_3^- + 10H^+ \rightarrow 4Zn^{2+} + NH_4^+ + 3H_2O \][/tex]
This is the balanced redox equation.
### (ii) Calculate the Percentage Purity of Zinc Sample
Given:
- Mass of impure zinc sample: [tex]\(4.0 \text{ g}\)[/tex]
- Volume of [tex]\( HNO_3 \)[/tex]: [tex]\( 25.0 \text{ mL} \)[/tex]
- Molarity of [tex]\( HNO_3 \)[/tex]: [tex]\(0.5 \text{ M}\)[/tex]
Step 1: Convert the volume of [tex]\( HNO_3 \)[/tex] from milliliters to liters.
[tex]\[ \text{Volume of } HNO_3 \text{ in L} = \frac{25.0 \text{ mL}}{1000} = 0.025 \text{ L} \][/tex]
Step 2: Calculate the moles of [tex]\( HNO_3 \)[/tex].
[tex]\[ \text{Moles of } HNO_3 = \text{Volume in L} \times \text{Molarity} = 0.025 \text{ L} \times 0.5 \text{ M} = 0.0125 \text{ mol} \][/tex]
Step 3: Determine the moles of [tex]\( Zn \)[/tex] that reacts.
From the balanced redox reaction:
[tex]\[ Zn + 2NO_3^- + 4H^+ \rightarrow Zn^{2+} + 2NO_2 + 2H_2O \][/tex]
We see that 1 mole of [tex]\( Zn \)[/tex] reacts with 2 moles of [tex]\( HNO_3 \)[/tex].
[tex]\[ \text{Moles of } Zn = \frac{\text{Moles of } HNO_3}{2} = \frac{0.0125}{2} = 0.00625 \text{ mol} \][/tex]
Step 4: Calculate the mass of pure [tex]\( Zn \)[/tex] that reacted.
[tex]\[ \text{Molar mass of } Zn = 65.38 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of pure } Zn = \text{Moles of } Zn \times \text{Molar mass} = 0.00625 \text{ mol} \times 65.38 \text{ g/mol} = 0.408625 \text{ g} \][/tex]
Step 5: Calculate the percentage purity of the zinc sample.
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure } Zn}{\text{Mass of impure sample}} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage purity} = \left( \frac{0.408625 \text{ g}}{4.0 \text{ g}} \right) \times 100 = 10.215625 \% \][/tex]
So, the percentage purity of the zinc sample is approximately [tex]\( 10.22 \% \)[/tex].
### (i) Balancing the Redox Equation Using Ion-Electron Method
First, let’s balance the given reaction using the ion-electron method. The reaction in question is:
[tex]\[ Zn + NO_3^- \rightarrow Zn^{2+} + NH_4^+ \][/tex]
Step 1: Write the oxidation and reduction half-reactions.
Oxidation half-reaction:
[tex]\[ Zn \rightarrow Zn^{2+} + 2e^- \][/tex]
Reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 2: Balance the atoms other than oxygen and hydrogen.
The zinc atoms in the oxidation half-reaction are already balanced.
Step 3: Balance oxygen atoms by adding [tex]\(H_2O\)[/tex] molecules.
In the reduction half-reaction:
There are no changes needed since it is already balanced for oxygen atoms.
Step 4: Balance hydrogen atoms by adding [tex]\(H^+\)[/tex] ions.
Again, the reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
The reduction half-reaction is balanced for hydrogen.
Step 5: Balance the charges by adding electrons.
The oxidation half-reaction:
[tex]\[ Zn \rightarrow Zn^{2+} + 2e^- \][/tex]
The reduction half-reaction:
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 6: Ensure the number of electrons lost in oxidation equals the number of electrons gained in reduction.
To balance the electrons, we need to multiply the oxidation half-reaction by 4 and the reduction half-reaction by 1:
[tex]\[ 4(Zn \rightarrow Zn^{2+} + 2e^-) \][/tex]
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Each side of the reaction should now have 8 electrons:
[tex]\[ 4Zn \rightarrow 4Zn^{2+} + 8e^- \][/tex]
[tex]\[ NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \][/tex]
Step 7: Combine the half-reactions and simplify.
Combining these, we get:
[tex]\[ 4Zn + NO_3^- + 10H^+ \rightarrow 4Zn^{2+} + NH_4^+ + 3H_2O \][/tex]
This is the balanced redox equation.
### (ii) Calculate the Percentage Purity of Zinc Sample
Given:
- Mass of impure zinc sample: [tex]\(4.0 \text{ g}\)[/tex]
- Volume of [tex]\( HNO_3 \)[/tex]: [tex]\( 25.0 \text{ mL} \)[/tex]
- Molarity of [tex]\( HNO_3 \)[/tex]: [tex]\(0.5 \text{ M}\)[/tex]
Step 1: Convert the volume of [tex]\( HNO_3 \)[/tex] from milliliters to liters.
[tex]\[ \text{Volume of } HNO_3 \text{ in L} = \frac{25.0 \text{ mL}}{1000} = 0.025 \text{ L} \][/tex]
Step 2: Calculate the moles of [tex]\( HNO_3 \)[/tex].
[tex]\[ \text{Moles of } HNO_3 = \text{Volume in L} \times \text{Molarity} = 0.025 \text{ L} \times 0.5 \text{ M} = 0.0125 \text{ mol} \][/tex]
Step 3: Determine the moles of [tex]\( Zn \)[/tex] that reacts.
From the balanced redox reaction:
[tex]\[ Zn + 2NO_3^- + 4H^+ \rightarrow Zn^{2+} + 2NO_2 + 2H_2O \][/tex]
We see that 1 mole of [tex]\( Zn \)[/tex] reacts with 2 moles of [tex]\( HNO_3 \)[/tex].
[tex]\[ \text{Moles of } Zn = \frac{\text{Moles of } HNO_3}{2} = \frac{0.0125}{2} = 0.00625 \text{ mol} \][/tex]
Step 4: Calculate the mass of pure [tex]\( Zn \)[/tex] that reacted.
[tex]\[ \text{Molar mass of } Zn = 65.38 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of pure } Zn = \text{Moles of } Zn \times \text{Molar mass} = 0.00625 \text{ mol} \times 65.38 \text{ g/mol} = 0.408625 \text{ g} \][/tex]
Step 5: Calculate the percentage purity of the zinc sample.
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure } Zn}{\text{Mass of impure sample}} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage purity} = \left( \frac{0.408625 \text{ g}}{4.0 \text{ g}} \right) \times 100 = 10.215625 \% \][/tex]
So, the percentage purity of the zinc sample is approximately [tex]\( 10.22 \% \)[/tex].
