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.


The half-reaction for the oxidation of the manganese in MnCO_3(s) to MnO_2(s).:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+


MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;

Step 1: Balance all the atom beside oxygen and hydrogen atom;

MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Manganese and carbon are balanced.

Step 2: Balance oxygen atom adding water on the required side:

MnCO_3+H_2O+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Step 3: Now balance hydrogen atom by adding hydrogen ion on the required side:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+

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

Explanation:The equation to be balanced is

MnCO3 > MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

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+

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