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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

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Answer:

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^+

Step-by-step explanation:


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^+

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