• Home
  • Chemistry
  • Astronomy
  • Energy
  • Nature
  • Biology
  • Physics
  • Electronics
  • Identifying the Reduced Element in the Reaction of 2HCl + 2KMnO4 → 3H2C2O4 + 6CO2 + 2MnO2 + 2KCl + 4H2O
    Here's how to determine the element that is reduced in the reaction:

    Understanding Oxidation and Reduction

    * Oxidation: Loss of electrons, increase in oxidation number.

    * Reduction: Gain of electrons, decrease in oxidation number.

    Analyzing the Reaction

    1. Assign Oxidation Numbers:

    * Reactants:

    * H in HCl: +1

    * Cl in HCl: -1

    * K in KMnO4: +1

    * Mn in KMnO4: +7

    * O in KMnO4: -2

    * H in H2C2O4: +1

    * C in H2C2O4: +3

    * O in H2C2O4: -2

    * Products:

    * C in CO2: +4

    * O in CO2: -2

    * Mn in MnO2: +4

    * O in MnO2: -2

    * K in KCl: +1

    * Cl in KCl: -1

    * H in H2O: +1

    * O in H2O: -2

    2. Identify Changes in Oxidation Numbers:

    * Manganese (Mn): Goes from +7 in KMnO4 to +4 in MnO2. This is a decrease in oxidation number, indicating reduction.

    * Carbon (C): Goes from +3 in H2C2O4 to +4 in CO2. This is an increase in oxidation number, indicating oxidation.

    Conclusion

    The element that is reduced in the reaction is Manganese (Mn).

    Science Discoveries © www.scienceaq.com