In analytical chemistry, the manganate ion actively undergoes a heavily documented, spontaneous disp — d and f Block Elements Chemistry Question
Question
In analytical chemistry, the manganate ion actively undergoes a heavily documented, spontaneous disproportionation chemical reaction when exposed to an acidic or strictly neutral medium, producing permanganate and solid manganese dioxide as dictated by the equation: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$. Determine the absolute numerical difference between the final oxidation state of Manganese in the distinctly oxidized product and its final oxidation state in the distinctly reduced product.
💡 Solution & Explanation
Initially, in the green manganate reactant ($MnO_4^{2-}$), Manganese possesses an oxidation state of $+6$. During disproportionation, it is pushed partially to an oxidized state in purple permanganate ($MnO_4^-$), which carries an oxidation state of $+7$. The remaining portion is heavily reduced to solid, black manganese dioxide ($MnO_2$), taking an oxidation state of $+4$. The absolute magnitude difference between these two terminating states is strictly $ | (+7) - (+4) | = 3$.