Question

he mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reaction for the oxidation of the manganese in MnCO3 to MnO2 in neutral groundwater where the principal carbonate species is HCO3–. Add H2O, H+, and electrons as needed to balance the half-reaction.

Answer 1

Answer:

$MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

Explanation:

Half reaction:$MnCO_{3}\rightarrow MnO_{2}$

(1)$CO_{3}$ balance: $MnCO_{3}\rightarrow MnO_{2}+HCO_{3}^{-}$

(2)H and O balance in acidic medium:$MnCO_{3}+2H_{2}O\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

(3) charge balance:$MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

Hence balanced half-reaction:$MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

Answer 2

Answer:

MnCO₃ + 2H₂O → MnO₂  + HCO₃⁻ + 2e- + 3H+

Explanation:

We have to first write out the chemical equation from the information we were given;

MnCO₃ → MnO₂  + HCO₃⁻

Like that, the reaction is unbalanced; in terms of atoms and charges.

It is observed that the oxidation number of Mn changes from +2 in MnCO₃  to +4 in MnO₂ , so two electrons must be added to the right. Since oxidation pretty much means loss of electrons.

The equation leads to;

MnCO₃  → MnO₂  + HCO₃⁻ + 2e-

In terms of electrons, the reaction is balanced.

However, in terms of charges it is not.

In the reactant side, total charge = 0.

In the product side, total charge = -1 + (-2) = -3

To balance this excess negative charge we add add 3 H+ on the product side;

MnCO₃ → MnO₂  + HCO₃⁻ + 2e- + 3H+

In terms of charges, the reaction is balanced.

However, in terms of atoms it is not.

We have excesses of Hydrogen and Oxygen atoms, to balance this we add H20 to the reactant side to balance the H and O atoms

This leads to;

MnCO₃ + 2H₂O → MnO₂  + HCO₃⁻ + 2e- + 3H+

The equation is now balanced.