Manganese ions (Mn^2+) can be oxidised by Persulphate ions S2O8^2- according to the following half-e — d and f Block Elements Chemistry Question
Question
Manganese ions (Mn^2+) can be oxidised by Persulphate ions S2O8^2- according to the following half-equations,<br>S2O8^2- + 2e^- $\rightarrow$ 2SO4^2-<br>Mn^2+ + 4H2O $\rightarrow$ MnO4^- + 8H^+ + 5e^-<br>How many moles of S2O8^2- are required to oxidise 1 mole of Mn^2+ ?
💡 Solution & Explanation
Step 1: Balance the electron transfer between the reduction of persulfate (gains 2e^- per mole) and the oxidation of manganese (loses 5e^- per mole). Step 2: To equate the electrons lost and gained, multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2. This gives: 5S2O8^2- + 10e^- ---> 10SO4^2- and 2Mn^2+ + 8H2O ---> 2MnO4^- + 16H^+ + 10e^-. Step 3: This shows that 5 moles of S2O8^2- are required to oxidize 2 moles of Mn^2+. Therefore, 2.5 moles of S2O8^2- are needed per 1 mole of Mn^2+, which corresponds to option (a).