Question

when the following equation is balanced properly under basic conditions, what are the coefficients of the species shown?
3 br₂ + 2 mno₂ → 6 br⁻ + 2 mno₄⁻
water appears in the balanced equation as a reactant (reactant, product, neither) with a coefficient of 4. (enter 0 for neither.)
how many electrons are transferred in this reaction? 6
an error has been detected in your answer. check for typos, miscalculations etc. before submitting your answer.
submit answer
retry entire group
9 more group attempts remaining

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

For oxidation: \(Br_{2}\to 2Br^{-}\) (balance \(Br\) atoms)
For reduction: \(MnO_{2}+2H_{2}O\to MnO_{4}^{-}+4H^{+}\) (balance \(O\) atoms using \(H_{2}O\) and \(H\) atoms using \(H^{+}\))

Step3: Balance charge in half - reactions (in basic medium, add \(OH^{-}\) to both sides to neutralize \(H^{+}\))

Oxidation: \(Br_{2}+2e^{-}\to 2Br^{-}\)
Reduction: \(MnO_{2}+4OH^{-}\to MnO_{4}^{-}+2H_{2}O + 2e^{-}\)

Step4: Make electron gain and loss equal

Multiply oxidation half - reaction by \(1\) and reduction half - reaction by \(1\)

Step5: Add half - reactions

\(Br_{2}+MnO_{2}+4OH^{-}\to 2Br^{-}+MnO_{4}^{-}+2H_{2}O\)

Step6: Check for electron transfer

From oxidation half - reaction (\(Br_{2}\to 2Br^{-}\)), \(Br\) changes from \(0\) to \(- 1\) per atom. For \(Br_{2}\) (2 \(Br\) atoms), total change in oxidation state is \(2\times(0 - (-1))=2\). From reduction half - reaction (\(MnO_{2}\to MnO_{4}^{-}\)), \(Mn\) changes from \(+4\) to \(+6\), change in oxidation state is \(+6-(+4) = 2\). Total electrons transferred \(= 2\)

Answer:

The coefficient of \(Br_{2}\) is \(1\), coefficient of \(MnO_{2}\) is \(1\), coefficient of \(Br^{-}\) is \(2\), coefficient of \(MnO_{4}^{-}\) is \(1\) and electrons transferred is \(2\).