Question

how many electrons must be added to the reactant side?
?e⁻ + 2h₂o + mno₄⁻ → mno₂ + 4oh⁻

Explanation:

Step1: Determine oxidation states

First, find the oxidation state of Mn in \(MnO_4^-\) and \(MnO_2\). In \(MnO_4^-\), let the oxidation state of Mn be \(x\). Oxygen has an oxidation state of -2, and the overall charge is -1. So, \(x + 4(-2)= -1\), which gives \(x - 8 = -1\), so \(x = +7\). In \(MnO_2\), let the oxidation state of Mn be \(y\). Then \(y + 2(-2)= 0\) (since \(MnO_2\) is neutral), so \(y - 4 = 0\), so \(y = +4\).

Step2: Calculate electron change

The oxidation state of Mn changes from +7 to +4, so each Mn atom gains \(7 - 4 = 3\) electrons. Now, check the number of Mn atoms: there is 1 Mn in \(MnO_4^-\) and 1 in \(MnO_2\), so the total number of electrons gained is 3. Therefore, we need to add 3 electrons to the reactant side to balance the charge (since electrons are gained, they are on the reactant side for reduction).

Answer:

3