Question

When the following oxidation-reduction equation representing a reaction that takes place in BASIC solution is correctly balanced using the smallest possible whole number coefficients, the coefficient before the H2O is : Al + MnO4- —-> A1(OH)4-+MnO2

Answer 1

Answer

The coefficient before the H₂O is D. 2

Explanation

Given that:

$Al+MnO_4^-\rightarrow Al(OH)_4^-+MnO_2$

What to find:

The coefficient before the H2O when the oxidation-reduction equation is correctly balanced in BASIC solution.

Step-by-step solution:

Step 1: Identify the species undergoing oxidation and reduction.

Al changes from 0 in Al to +3 in Al(OH)₄⁻. It is undergoing oxidation.

Mn changes from +7 in MnO)₄⁻ to +4 in MnO₂. It is undergoing a reduction.

Step 2: Write the half-reaction for Oxidation and reduction.

$\begin{gathered} Oxidation:Al\rightarrow Al(OH)_4^-+3e^- \\ \\ Reduction:MnO_4^-+3e^-\rightarrow MnO_2 \end{gathered}$

Since the number of electrons is equal, this will be canceled so the next step is to write the overall reaction.

Step 3: Write the overall reaction.

$Al+MnO_4^-\rightarrow Al(OH)_4^-+MnO_2$

Step 4: Balance the number of Oxygen atoms by adding H₂O on the left side.

$Al+MnO_4^-+2H_2O\rightarrow Al(OH)_4^-+MnO_2$

The overall balanced equation using the smallest possible whole-number coefficients is:

$Al+MnO_4^-+2H_2O\rightarrow Al(OH)_4^-+MnO_2$

Therefore, the coefficient before the H₂O is 2