Question

a student is asked to balance an equation by using the half - reaction method. he determines the two half reactions as shown below.

$2\ce{br^-} \longrightarrow \ce{br_2} + 2\ce{e^-}$

$\ce{cl_2} + 2\ce{e^-} \longrightarrow 2\ce{cl^-}$

what should he write as the final, balanced equation?

$\circ$ $\ce{cl_2} + 2\ce{br^-} \longrightarrow \ce{br_2} + 2\ce{cl^-}$
$\circ$ $\ce{cl_2} + 2\ce{br^-} + 2\ce{e^-} \longrightarrow \ce{br_2} + 2\ce{cl^-}$
$\circ$ $\ce{cl_2} + 2\ce{br^-} \longrightarrow \ce{br_2} + 2\ce{cl^-} + 2\ce{e^-}$
$\circ$ $\ce{cl_2} + \ce{br^-} + 2\ce{e^-} \longrightarrow \ce{br_2} + \ce{cl^-} + 2\ce{e^-}$

Explanation:

Step1: Recall half - reaction combination

To balance a redox reaction using half - reactions, we add the two half - reactions together. The first half - reaction is $\ce{2Br^{-} -> Br_{2} + 2e^{-}}$ (oxidation, loss of electrons) and the second is $\ce{Cl_{2} + 2e^{-} -> 2Cl^{-}}$ (reduction, gain of electrons).

Step2: Add the half - reactions

When we add the two half - reactions, the electrons ($\ce{2e^{-}}$) on the product side of the first reaction and the reactant side of the second reaction will cancel out. So we add $\ce{2Br^{-} -> Br_{2} + 2e^{-}}$ and $\ce{Cl_{2} + 2e^{-} -> 2Cl^{-}}$:
$\ce{2Br^{-} + Cl_{2} + 2e^{-} -> Br_{2} + 2Cl^{-} + 2e^{-}}$
After canceling the electrons, we get $\ce{Cl_{2} + 2Br^{-} -> Br_{2} + 2Cl^{-}}$.

Answer:

A. $\ce{Cl_{2} + 2Br^{-} -> Br_{2} + 2Cl^{-}}$