Question

the electrons are on the reactants side.
?e⁻ + 4h⁺ + mno₂ → mn²⁺ + 2h₂o
how many electrons are required to balance the charge in the half - reaction?

Explanation:

Step1: Calculate charge on left side

The left side has \(4\text{H}^+\) (charge \(+4\)) and \(\text{MnO}_2\) (charge \(0\) for the compound, but we consider the charge contribution from ions; \(\text{MnO}_2\) is a neutral compound here, and we also have \(x\) electrons (\(e^-\)) with charge \(-x\)). So total left charge: \(+4 - x\).

Step2: Calculate charge on right side

The right side has \(\text{Mn}^{2+}\) (charge \(+2\)) and \(2\text{H}_2\text{O}\) (charge \(0\) for each water molecule, so total \(0\)). So total right charge: \(+2\).

Step3: Set charges equal and solve for x

For charge balance, left charge = right charge:
\(+4 - x = +2\)
Subtract \(4\) from both sides: \(-x = +2 - 4 = -2\)
Multiply both sides by \(-1\): \(x = 2\).

Answer:

2