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

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: $2KI + Br_2 \rightarrow 2KBr + I_2$
• 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: $2I^- \rightarrow I_2 + 2e^-$
• Bromine (Br) in Br2 is reduced from Br2 (oxidation state 0) to Br^- in KBr: $Br_2 + 2e^- \rightarrow 2Br^-$

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:
  $2I^- \rightarrow I_2 + 2e^-$

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