# Which of the following reactions is/are spontaneous? (i) Cl_2 + 2Br^(-) -> Br_2 + 2Cl^(-) (ii) Br_2 + 2I^(-) -> I_2 + 2Br^(-)

Jul 12, 2015

Both of those reactions are spontaneous.

#### Explanation:

You're actually dealing with two redox reactions, which means that you can easily figure out which one, if any, is spontaneous by looking at the standard reduction potentials for the half-reactions.

Take the first reaction

$C {l}_{2 \left(g\right)} + 2 B {r}_{\left(a q\right)}^{-} \to B {r}_{2 \left(l\right)} + 2 C {l}_{\left(a q\right)}^{-}$

The standard reduction potentials for the half-reactions are

$B {r}_{2 \left(l\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 B {r}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+1.09 V}$

$C {l}_{2 \left(g\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 C {l}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+1.36 V}$

In order for the reaction to take place, you need chlorine to oxidize the bromide anion to liquid bromime, and be reduced to the chloride anion in the process.

Since chlorine has a more positive ${E}^{\circ}$ value, it will be more than cpable of doing just that. This means that the first equilibrium reaction will actually move to the left, and the second equilibrium reaction will move to the right.

The standard cell potential for the overall reaction will thus be

${E}_{\text{cell"^@ = E_"cathode"^@ + E_"anode}}^{\circ}$

${E}_{\text{cell"^@ = "1.36 V" +underbrace((- "1.09 V"))_(color(blue)("because the equilibrium moves to the left!")) = "+0.27 V}}$

The spontaneity of the cell is given by the equation

$\Delta {G}^{\circ} = - n F \cdot {E}_{\text{cell}}^{\circ}$, where

$n$ - the number of electrons exchanged in the reaction;
$F$ - Faraday's constant.

This basically tells you that, in order for the cell reaction to be spontaneous, $\Delta {G}^{\circ}$ must be negative, which implies that ${E}_{\text{cell}}^{\circ}$ must be positive.

Since this is the case for the first reaction, it is indeed spontaneous.

The same approach can be used for the second reaction.

$B {r}_{2 \left(l\right)} + 2 {I}_{\left(a q\right)}^{-} \to {I}_{2 \left(a q\right)} + 2 B {r}_{\left(a q\right)}^{-}$

Once again, use the standard electrode potentials

${I}_{2 \left(s\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 {I}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+0.54 V}$

$B {r}_{2 \left(l\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 B {r}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+1.09 V}$

This time, you need bromine to oxidize the iodide anion to iodine, and be reduced in the process. The more positive ${E}^{\circ}$ value for bromine's reduction half-reaction confirms that this is what's going to happen.

The first equilibrium will again shift to the left, and the second equilibrium to the right. This means that you have

${E}_{\text{cell"^@ = E_"cathode"^@ + E_"anode}}^{\circ}$

${E}_{\text{cell"^@ = "+1.09 V" + underbrace((-"0.54 V"))_(color(blue)("because the equilibrium shifts left!")) = "+0.55 V}}$

Again, a positive ${E}_{\text{cell}}^{\circ}$ implies a negative $\Delta {G}^{\circ}$, and thus a spontaneous reaction.