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Question 1:

What is the most accurate answer when we balance the following reaction using Half-Reaction for Acidic solution?

MnO4^- + 2CHCl3 → 3Cl2 + 2CO2 + Mn^2+

For the Final answer:

Question 1 options:
a. We have 4H2O on the product side
b. We have 6H^+ on the products side
c. We have 6H2O on the reactants side
d. We have 4H^+ on the reactants side

5 answers

To balance the given redox reaction using the half-reaction method for an acidic solution, let’s break it down step-by-step.

  1. Identify the half-reactions:

    • The reduction half-reaction involves \( \text{MnO}_4^- \) reducing to \( \text{Mn}^{2+} \).
    • The oxidation half-reaction involves \( \text{CHCl}_3 \) oxidizing to \( \text{Cl}_2 \) and \( \text{CO}_2 \).

  2. Balance the reduction half-reaction: The reduction half-reaction: \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

  3. Balance the oxidation half-reaction: The oxidation of \( \text{CHCl}_3 \) to \( \text{Cl}_2 \): \[ 2 \text{CHCl}_3 \rightarrow 3 \text{Cl}_2 + 2 \text{CO}_2 + 6 \text{H}^+ + 6 \text{e}^- \]

  4. Combine the half-reactions: To combine these, we need to ensure the number of electrons lost equals the number gained. We have 5 electrons gained in the reduction and 6 electrons lost in the oxidation. The least common multiple of 5 and 6 is 30. Therefore, we need to multiply the half-reactions accordingly:

    • The reduction reaction must be multiplied by 6: \[ 6 \text{MnO}_4^- + 48 \text{H}^+ + 30 \text{e}^- \rightarrow 6 \text{Mn}^{2+} + 24 \text{H}_2\text{O} \]
    • The oxidation reaction must be multiplied by 5: \[ 10 \text{CHCl}_3 \rightarrow 15 \text{Cl}_2 + 10 \text{CO}_2 + 30 \text{H}^+ + 30 \text{e}^- \]

  5. Final balanced equation: Combining these, we get: \[ 6 \text{MnO}_4^- + 10 \text{CHCl}_3 + 18 \text{H}^+ \rightarrow 6 \text{Mn}^{2+} + 15 \text{Cl}_2 + 10 \text{CO}_2 + 24 \text{H}_2\text{O} \]

From this balanced equation, we can analyze the options provided:

  • Option a (4H2O on product side): This is incorrect (we have 24 H2O).
  • Option b (6H^+ on the products side): This is incorrect (we have 18 H⁺ on the reactants side).
  • Option c (6H2O on reactants side): This is incorrect (we have 24 H2O on the product side).
  • Option d (4H^+ on the reactants side): This is incorrect (we have 18 H⁺ on the reactants side).

Ultimately: None of the given options seem correct because all descriptions from the options do not match the balanced reaction. Please verify the question or the options provided again.

are you sure?

Let's carefully review the balanced half-reaction for the redox reaction one more time to clarify the issue with the options given.

Given Reaction

The reaction we need to balance is: \[ \text{MnO}_4^- + 2\text{CHCl}_3 \rightarrow 3\text{Cl}_2 + 2\text{CO}_2 + \text{Mn}^{2+} \]

Step 1: Write Half-Reactions

  1. Reduction Half-Reaction: \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

  2. Oxidation Half-Reaction:

    • The oxidation involves converting \( \text{CHCl}_3 \) to \( \text{Cl}_2 \) and producing \( \text{CO}_2 \): \[ 2 \text{CHCl}_3 \rightarrow 3 \text{Cl}_2 + 2 \text{CO}_2 + 6 \text{H}^+ + 6 \text{e}^- \]

Step 2: Adjust Half-Reactions

To balance the number of electrons:

  • Multiply the reduction half-reaction by 6: \[ 6 \text{MnO}_4^- + 48 \text{H}^+ + 30 \text{e}^- \rightarrow 6 \text{Mn}^{2+} + 24 \text{H}_2\text{O} \]
  • Multiply the oxidation half-reaction by 5: \[ 10 \text{CHCl}_3 \rightarrow 15 \text{Cl}_2 + 10 \text{CO}_2 + 30 \text{H}^+ + 30 \text{e}^- \]

Final Balanced Equation

Combining these gives: \[ 6 \text{MnO}_4^- + 10 \text{CHCl}_3 + 18 \text{H}^+ \rightarrow 6 \text{Mn}^{2+} + 15 \text{Cl}_2 + 10 \text{CO}_2 + 24 \text{H}_2\text{O} \]

Analysis of the Options

Now we analyze the options provided:

a. We have 4H2O on the product side: Incorrect (we have 24 H₂O).

b. We have 6H^+ on the products side: Incorrect (we have 18 H⁺ on the reactant side).

c. We have 6H2O on the reactants side: Incorrect (we have 24 H₂O on the product side).

d. We have 4H^+ on the reactants side: Incorrect (we have 18 H⁺ on the reactants side).

Conclusion:

None of the options appear to be accurate based on the balanced equation derived from the half-reaction method. If the options were supposed to describe a correct balanced reaction, they should relate to the actual counts in the final balanced equation. This suggests either a misunderstanding in the question or potential misstatements in the options themselves.

If you have the original problem statement or context, please provide it so I can assist you more effectively!

is it d)?

Let's revisit the balanced half-reaction process to finally clarify the correct answer.

Given Reaction

The initial reaction to balance is: \[ \text{MnO}_4^- + 2\text{CHCl}_3 \rightarrow 3\text{Cl}_2 + 2\text{CO}_2 + \text{Mn}^{2+} \]

Half-Reactions

  1. Reduction Half-Reaction for \(\text{MnO}_4^-\): \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

  2. Oxidation Half-Reaction for converting \(\text{CHCl}_3\) to \(\text{Cl}_2\): \[ 2 \text{CHCl}_3 \rightarrow 3 \text{Cl}_2 + 2 \text{CO}_2 + 6 \text{H}^+ + 6 \text{e}^- \]

Combine the Half-Reactions

To balance the electrons:

  • Multiply the reduction half-reaction by 6: \[ 6 \text{MnO}_4^- + 48 \text{H}^+ + 30 \text{e}^- \rightarrow 6 \text{Mn}^{2+} + 24 \text{H}_2\text{O} \]

  • Multiply the oxidation half-reaction by 5: \[ 10 \text{CHCl}_3 \rightarrow 15 \text{Cl}_2 + 10 \text{CO}_2 + 30 \text{H}^+ + 30 \text{e}^- \]

Final Balanced Chemical Equation

Combining everything we get: \[ 6 \text{MnO}_4^- + 10 \text{CHCl}_3 + 18 \text{H}^+ \rightarrow 6 \text{Mn}^{2+} + 15 \text{Cl}_2 + 10 \text{CO}_2 + 24 \text{H}_2\text{O} \]

Analyzing the Options:

  • Option a (4H2O on the product side): Incorrect (we have 24 H₂O).
  • Option b (6H^+ on the products side): Incorrect (we have 18 H⁺ on the reactants side).
  • Option c (6H2O on the reactants side): Incorrect (we have 24 H₂O on the product side).
  • Option d (4H^+ on the reactants side): Incorrect (we actually have 18 H⁺ on the reactants side).

Conclusion

None of the options listed are correct since they all misstate the balancing outcome achieved. The balanced reaction specifies 18 H⁺ ions on the reactants side rather than any of the options offered.

If you were considering "d" (which states that we have 4 H⁺ on the reactants side), it is also not correct based on the balanced equation.

If there's any confusion or a potential mix-up, please let me know, and I will gladly help clarify further!