Which of the following is the cell reaction that occurs when the following half-cells are combined?
`I_2 + 2e^(-) to 2I^(-) (1M) , E^@=+0.54 V`
`Br_2+2e^(-) to 2Br (1 M) , E^@=+1.09 V`

Options

A. `2Br^(-) + I_2 to Br_2 + 2I^-`
B. `I_2 + Br_2 to 2I^(-) + 2Br^-`
C. `2I^(-) + Br_2 to I_2 + 2Br^-`
D. `2I^(-) + 2Br^(-) to I_2 + Br_2`

Explanation

To determine the overall cell reaction when the given half-cells are combined, we need to follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
- The first half-reaction is:
\[ I_2 + 2e^- \rightarrow 2I^- \]
with a standard reduction potential \( E^\circ = +0.54 \, \text{V} \).

- The second half-reaction is:
\[ Br_2 + 2e^- \rightarrow 2Br^- \]
with a standard reduction potential \( E^\circ = +1.09 \, \text{V} \).

### Step 2: Determine which half-reaction will occur at the cathode and which at the anode.
- The half-reaction with the higher reduction potential will occur at the cathode (reduction), while the one with the lower reduction potential will occur at the anode (oxidation).
- Since \( E^\circ_{Br_2} = +1.09 \, \text{V} \) is greater than \( E^\circ_{I_2} = +0.54 \, \text{V} \), \( Br_2 \) will be reduced at the cathode, and \( I_2 \) will be oxidized at the anode.

### Step 3: Write the oxidation half-reaction for the anode.
- The oxidation of \( I^- \) can be represented as:
\[ 2I^- \rightarrow I_2 + 2e^- \]

### Step 4: Write the reduction half-reaction for the cathode.
- The reduction of \( Br_2 \) can be represented as:
\[ Br_2 + 2e^- \rightarrow 2Br^- \]

### Step 5: Combine the half-reactions to get the overall cell reaction.
- The overall cell reaction is obtained by adding the oxidation and reduction half-reactions:
\[ 2I^- + Br_2 \rightarrow I_2 + 2Br^- \]

### Final Answer:
The overall cell reaction is:
\[ 2I^- + Br_2 \rightarrow I_2 + 2Br^- \]