Question

when the following equation is balanced properly under basic conditions, what are the coefficients of the species shown?
br₂ + mno₂ → br⁻ + mno₄⁻
water appears in the balanced equation as a (reactant, product, neither) with a coefficient of. (enter 0 for neither.)
how many electrons are transferred in this reaction?

Explanation:

Step1: Write half - reactions

Oxidation half - reaction: \(Br_{2}\to Br^{-}\)
Reduction half - reaction: \(MnO_{2}\to MnO_{4}^{-}\)

Step2: Balance atoms in half - reactions

Oxidation: \(Br_{2}\to 2Br^{-}\) (balance Br atoms)
Reduction: \(MnO_{2}+2H_{2}O\to MnO_{4}^{-}+4H^{+}\) (balance O atoms with \(H_{2}O\) and H atoms with \(H^{+}\))

Since it is basic medium, add \(OH^{-}\) to both sides to neutralize \(H^{+}\)
Reduction: \(MnO_{2}+4OH^{-}\to MnO_{4}^{-}+2H_{2}O + 2e^{-}\)

Oxidation: \(Br_{2}+2e^{-}\to 2Br^{-}\)

Step3: Balance electrons and combine half - reactions

Multiply oxidation reaction by 1 and reduction reaction by 1 (electrons are already balanced, 2 electrons in each)
\(Br_{2}+MnO_{2}+4OH^{-}\to 2Br^{-}+MnO_{4}^{-}+2H_{2}O\)

Step4: Determine coefficient of \(H_{2}O\)

From the balanced equation \(Br_{2}+MnO_{2}+4OH^{-}\to 2Br^{-}+MnO_{4}^{-}+2H_{2}O\), the coefficient of \(H_{2}O\) is 2

Step5: Calculate number of electrons transferred

In oxidation half - reaction \(Br_{2}\to 2Br^{-}\), each Br atom changes from 0 to - 1. For \(Br_{2}\) (2 Br atoms), total change in oxidation state is \(2\times(0 - (- 1))=2\)
In reduction half - reaction \(MnO_{2}\to MnO_{4}^{-}\), Mn changes from + 4 to + 6 (change of + 2). But since there is 1 Mn atom, and considering electron transfer balance (from step 3 where 2 electrons are transferred in each half - reaction), the number of electrons transferred in the overall reaction is 2

Answer:

The coefficient of \(H_{2}O\) is 2 and the number of electrons transferred is 2.