Write the balanced half-reaction for the conversion of \(\operatorname{HOBr}(a q)\) to \(\mathrm{Br}_{2}(a q)\) in acidic solution.

Short Answer

Expert verified
The balanced half-reaction for the conversion of \(\operatorname{HOBr}(a q)\) to \(\mathrm{Br}_{2}(a q)\) in acidic solution is: \( 2\operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) + O_{2}H_{2} + 2H^{+}(a q) + 6e^{-} \)

Step by step solution

01

Balance Atoms Other Than Hydrogen and Oxygen

The balanced skeletal equation for the conversion is: \( \operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) \) Here, Bromine (Br) atoms are not balanced. To balance the bromine, we add coefficient of 2 to \( \operatorname{HOBr} \) which leads to: \( 2\operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) \)
02

Balance Oxygen Atoms by Adding Water Molecules

Now balance the oxygen atoms by adding water molecules to the right side of equation: \( 2\operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) + O_{2}H_{2} \)
03

Balance Hydrogen Atoms and Charge

The hydrogen atoms can now be balanced by adding hydrogen ions to the right side of the equation: \( 2\operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) + O_{2}H_{2} + 2H^{+}(a q) \) Furthermore, to balance the charge we add six electrons to the right side: \( 2\operatorname{HOBr}(a q) \rightarrow \mathrm{Br}_{2}(a q) + O_{2}H_{2} + 2H^{+}(a q) + 6e^{-} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free