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In a particular redox reaction, BrO− is oxidized to BrO−3 and Ag + is reduced to Ag.
Complete and balance the equation for this reaction in acidic solution. Phases are optional.
BrO ⁻ + Ag ⁺ → BrO₃ ⁻ + Ag

Answer :

dsdrajlin

Answer:

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺

Explanation:

In order to balance a redox reaction, we will use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Ag⁺ → Ag

Oxidation: BrO⁻ → BrO₃⁻

Step 2: Perform the mass balance adding H⁺ and H₂O where necessary

Ag⁺ → Ag

2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺

Step 3: Perform the electrical balance adding electrons where necessary

1 e⁻ + Ag⁺ → Ag

2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺ + 4 e⁻

Step 4: Multiply both half-reactions by numbers that secure that the number of electrons gained and lost are equal

4 × (1 e⁻ + Ag⁺ → Ag)

1 × (2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺ + 4 e⁻)

Step 5: Add both half-reactions

4 e⁻ + 4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺ + 4 e⁻

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺

Histrionicus

The complete and balanced equation for redox reaction, BrO− is oxidized to BrO−³ and Ag + is reduced to Ag in acidic solution is -

  • 4Ag⁺ + BrO⁻+ 2H₂O ----> 4 Ag + BrO₃⁻+ 4H⁺

The steps or phases for this redox or reduction-oxidation reactions:

  • first separate the equation into the two reactions that are happening BrO⁻ ----> BrO3⁻ and Ag+ ----> Ag
  • Now Balance them one at a time

Start with BrO

  • the first step is always to balance the O. We do this by adding water.

BrO⁻ + 2H₂O ----> BrO₃⁻

  • The next step is to balance the H. We do this by adding H+

BrO⁻+ 2H₂O ----> BrO₃⁻+ 4H⁺

  • Next we balance the charge, by adding electrons.
  1. The charge on the left side comes only from BrO so the charge is -1.
  2. The charge on the right side is -1 from BrO₃ +4 from H+, so +3. To make the charge equal on both sides we need to add 4 electrons to the side with the +3 charge, then it will be -1.

BrO⁻ + 2 H₂0 ----> BrO₃ ^- + 4H⁺ + 4e⁻

This equation is now balanced. Now to the Ag.

Ag⁺ ---> Ag

  • The only thing that changed was the charge and can be corrected by adding electrons.

Ag⁺ + e⁻ ----> Ag

The charge is now balanced.

  • Now when combine these two half reactions, we want the electrons to cancel out.
  • The reaction required 4 electrons in the Ag equation to balance the 4e need in the BrO equation.
  • So we multiply the entire Ag equation by 4

4Ag⁺ + 4e⁻ ------> 4Ag

Thus, at last combine both Ag equation and BrO equation and cancel out our electrons and get

- 4Ag⁺ + BrO⁻+ 2H₂O ----> 4 Ag + BrO₃⁻+ 4H⁺

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