Find the correct stoichiometric coefficients in the following redox equation:
\(xH^+ + yC_6H_5CH_3 + wMnO_4^- \longrightarrow aH_2O + bMn^{2+} + tC_6H_5COOH\)
Answer & explanation
Correct answer: option 1
The correct answer is option 1.
| x | y | w | a | b | t |
| 18 | 5 | 6 | 14 | 6 | 5 |
Let us break down the redox equation and the process of balancing it:
The given redox equation is:
\[ xH^+ + yC_6H_5CH_3 + wMnO_4^- \longrightarrow aH_2O + bMn^{2+} + tC_6H_5COOH \]
To balance this equation, we need to first identify the elements undergoing oxidation and reduction. In this case:
1. Oxidation half-reaction: \( C_6H_5CH_3 \rightarrow C_6H_5COOH \)
2. Reduction half-reaction: \( MnO_4^- \rightarrow Mn^{2+} \)
Now, let's balance each half-reaction:
1. Oxidation half-reaction: The oxidation state of carbon in methylbenzene (\(C_6H_5CH_3\)) changes from 0 to +2 in benzoic acid (\(C_6H_5COOH\)). So, 2 electrons are involved in the oxidation half-reaction.
\[ 5C_6H_5CH_3 + 14H^+ + 14e^- \rightarrow 5C_6H_5COOH + 14H_2O \]
2. Reduction half-reaction: The oxidation state of manganese changes from +7 to +2. To balance the charges and atoms, we need 12 electrons.
\[ 6MnO_4^- + 18H^+ + 12e^- \rightarrow 6Mn^{2+} + 10H_2O \]
Now, let's combine the half-reactions and balance the coefficients:
\[ 18H^+ + 5C_6H_5CH_3 + 6MnO_4^- \rightarrow 14H_2O + 6Mn^{2+} + 5C_6H_5COOH \]
Comparing the coefficients, we find that they match the coefficients:
\[ x = 18, \quad y = 5, \quad w = 6, \quad a = 14, \quad b = 6, \quad t = 5 \]
Therefore, option 1 is the correct choice for the stoichiometric coefficients that balance the given redox equation.