MnSO4 is converted to MnO2 in a basic medium as oxygen gas is converted to hyrox
ID: 754082 • Letter: M
Question
MnSO4 is converted to MnO2 in a basic medium as oxygen gas is converted to hyroxide ion. In this case, is the sulfate ion a spectator ion? (NaOH is also present. Is sodium ion also a spectator ion?) A. Print the oxidation half-equation B. Print the reduction half-equation C. Print the Net redox ionic equation (without any spectator ions unless they are involved in the balance) Please include explanation!Explanation / Answer
Manganese changes its oxidation state from +4 in MnO2 to +2 in MnSO4, whereas carbon changes its ox state from +3 in H2C2O4 to +4 in CO2. Write down half reactions now: MnO2 +2e --> Mn(2+) To remove oxygen from MnO2 (since there is no oxygen attached to Mn in the right side of the quation), you need to add either H2O or H(+) to the left side and add OH(-) or H2O to the right side, respectively. You have acidic medium, so add H(+): MnO2 + 4H(+) + 2e --> Mn(2+) + 2H2O That is first complete half-reaction. Similarly, H2C2O4 - 2e --> CO2 + 2H(+) . I added two H(+) to the right top balance hydrogens. This is the second half-reaction. Combine two half-reactions together (conviniently, you have the same number of electrons removed as accepted, so you do not have to multiply any half-reaction by any number to balance electrons): MnO2 + 4H(+) + H2C2O4 --> Mn(2+) + CO2 + 2H(+) + 2H2O or MnO2 + 2H(+) + H2C2O4 --> Mn(2+) + CO2 + 2H2O Now, in molecular form (you know that H(+) come from H2SO4): MnO4 + H2SO4 + H2C2O4 --> MnSO4 + CO2 + 2H2O