The following chemical reaction occurs in a basic solution. mg2+(aq) + mno2(aq) + oh−(aq) → mg(s) + mno4−(aq) + h2o(l) how many moles of electrons are transferred when the equation is balanced using the smallest whole-number coefficients? group of answer choices

6 moles of electrons

Explanation:

Firstly, separate the ionic equation into reduction and oxidation half equations. Before doing this, note the following:

I.Oxidation is the loss of electron(s) while reduction is the gain of electron(s).

II.The oxidizing agent undergoes reduction while the reducing agent undergoes oxidation.

III.The oxidation number of the oxidizing agent decreases in the positive direction while that of the reducing agent increases in the positive direction.

IV.The electron gain for reduction is written on the left hand side of the equation arrow ( with a plus sign ) while the electron loss for oxidation is written on the right hand side of the equation arrow ( with a plus sign ).

For the above question, the oxidation number of Mg changed from +2 to zero . Since it decreased, Mg2+ is the oxidizing agent and undergoes reduction

I.Mg2+(aq) + 2e = Mg(s) { reduction}

II.MnO2 + OH- = MnO4- + H2O

First, balance the atoms

MnO2 + 4OH- = MnO4- + 2H2O

Checking the number of charges on both equation sides we have -4 on the left hand side and -1 on the right hand side. Since oxidation number increased from -4 to -1, add the 3e loss to the right hand side

III.MnO2 + 4OH- = MnO4- + 2H2O + 3e { oxidation }

But redox reactions are opposing yet complementary processes, meaning that the electron loss for oxidation must be equal to the electron gain for reduction.

To achieve this, multiply the initial reduction half equation (I) by 3 and the initial oxidation half equation (III) by 2 to obtain the accurate reduction and oxidation half reactions as:

IV.3Mg2+(aq) + 6e = 3Mg(s) { reduction }

V.2MnO2(aq) + 8OH-(aq) = 2MnO4-(aq) + 4H2O(l) + 6e { oxidation }

From the balanced half reactions it is evident that 6 moles of electrons were transferred.

The balanced ionic equation for the above reaction is:

3Mg2+(aq) + 2MnO2(aq) + 8OH-(aq) = 3Mg(s) + 2MnO4-(aq) + 4H2O(l).

You can check if your equation is balanced noting that the total number of charges on the left hand side must be equal to that on the right hand side.

Left hand side charges:

3(+2) + 8(-1) = +6-8 = -2

Right hand side charges:

2(-1) = -2.