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Write the oxidation and reduction half-reactions (without states of matter) for the following reaction:→+Br2aq  2KIaq + 2KBraq  I2aq

Question

Write the oxidation and reduction half-reactions (without states of matter) for the following reaction:

Br2+2KI2KBr+I2 \text{Br}_2 + 2 \text{KI} \rightarrow 2 \text{KBr} + \text{I}_2

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Solution

To identify the oxidation and reduction half-reactions for the given reaction, we need to recognize the changes in oxidation states of the elements involved.

1. Break Down the Problem

  • The reaction can be summarized as follows: 2KI+Br22KBr+I2 2KI + Br_2 \rightarrow 2KBr + I_2
  • We will separate this into two half-reactions: one for oxidation and one for reduction.

2. Relevant Concepts

  • Oxidation is the loss of electrons or an increase in oxidation state.
  • Reduction is the gain of electrons or a decrease in oxidation state.

3. Analysis and Detail

  • In the reaction, iodine (I) in KI is oxidized to I2 (oxidation state changes from -1 in KI to 0 in I2), thus it undergoes oxidation: 2II2+2e 2I^- \rightarrow I_2 + 2e^-
  • Bromine (Br) in Br2 is reduced from Br2 (oxidation state 0) to Br^- in KBr: Br2+2e2Br Br_2 + 2e^- \rightarrow 2Br^-

4. Verify and Summarize

  • Both half-reactions are balanced concerning the number of atoms and charge. The oxidation half-reaction accounts for 2 electrons being released, which matches with the reduction half-reaction that consumes 2 electrons.

Final Answer

  • Oxidation Half-Reaction:
    2II2+2e 2I^- \rightarrow I_2 + 2e^-

  • Reduction Half-Reaction:
    Br2+2e2Br Br_2 + 2e^- \rightarrow 2Br^-

This problem has been solved

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