Question #b2056

Apr 8, 2017

The Chlorine is the oxidizing agent. The Bromine is the reducing agent.

Explanation:

$2 N {a}^{+} B {r}^{-} 1 + C {l}_{2}^{o} = 2 N {a}^{+} C {l}^{-} + B {r}_{2}^{o}$

Cl goes from a zero charge to a -1 charge The chlorine has been reduced its charge has gone down.

Br goes from -1 to a zero charge. The Bromine has been oxidized it charge has gone up

Apr 8, 2017

$N a B r \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N a C l \left(s\right) + \frac{1}{2} B {r}_{2} \left(l\right)$, and this is a stoichiometrically balanced equation.

Explanation:

And if we assign oxidation numbers, we can immediately see the direction of electron transfer:

$\stackrel{+ I}{N a} \stackrel{- I}{B r} \left(s\right) + \frac{1}{2} \stackrel{0}{C} {l}_{2} \left(g\right) \rightarrow \stackrel{+ I}{N a} \stackrel{- I}{C l} \left(s\right) + \frac{1}{2} \stackrel{0}{B} {r}_{2}$

Bromide ion has been oxidized; chlorine gas has been reduced. Note (i) that you simply must know that the halogens are normally bimolecular; we represent them as ${X}_{2}$.

Also note (ii) that we probably could not get this reaction to work, even if thermodynamics is on our side.