Question

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

Answer 1

Answer:

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

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

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:

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