Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-containing species in the product is HCO3–(aq). Add H2O and H+ to balance the H and O atoms in the equation. Do not add electrons; you may leave the half-reaction unbalanced with respect to charge.

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

shrutiagrawal1798 shrutiagrawal1798 · Answer 2 · 2022-01-14T12:54:28+01:00

The half reaction for the oxidation of Manganese has been [tex]\rm MnCO_3\;+\;2\;H_2O\;\rightarrow\;MnO_2\;+\;HCO_3^-\;+\;2e^-\;+\;3\;H^+[/tex]

Oxidation has been defined as the loss of electrons by an element. The half reaction has been the oxidation reaction in the redox reaction.

Half reaction of oxidation

In the reaction of Manganese carbonate to manganese dioxide, there has been change in the oxidation number of Mn from +2 to +4. The change in the oxidation state has been balanced with the addition of electrons.

The carbonate charge has been balanced in the reaction with the addition of hydrogen and water molecules.

The half reaction for the oxidation of Manganese has been [tex]\rm MnCO_3\;+\;2\;H_2O\;\rightarrow\;MnO_2\;+\;HCO_3^-\;+\;2e^-\;+\;3\;H^+[/tex]

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