Question

Question-98: A potential difference of 2.0V is impressed across a pair of inert electrodes (e.g., Platinum) that are immersed in a 0.05M aqueous KBr solution. What are the products that form at the anode and the cathode?

Short answer

The products that form at the anode and the cathode are : Br₂ and K respectively.

At cathode:

${\text{K}}^{\text{+}}{\text{(aq) +e}}^{\text{-}}\to \text{K(s)}$

At anode:

$2B{r}^{-}\left(aq\right)+2e^{-}\to Br2\left(l\right)$

Step-by-step solution

Step-1: Chemical reaction at cathode and anode

At cathode:

$$\begin{gathered} {\text{K}}^{\text{+}}{\text{(0.05M) +e}}^{\text{-}}\to \text{K} \\ {\text{2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(10}}^{\text{- 7}}{\text{M) + 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O} \end{gathered}$$

At anode:

$$\begin{gathered} {\text{2Br}}^{\text{-}}\to {\text{Br}}_{\text{2}}{\text{+ 2e}}^{\text{-}} \\ {\text{6H}}_{\text{2}}\text{O}\to {\text{O}}_{\text{2}}{\text{+ 4H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(10}}^{\text{- 7}}{\text{M) + 4e}}^{\text{-}} \end{gathered}$$

Step-2: Reduction potential of first half-reaction

Using following method, we can calculate reduction potential:

$\begin{array}{rcl}& & \text{E}\left(K^{+}/K\right){\text{=E}}^{\text{o}}\left(K^{+}/K\right)\text{-}\frac{\text{0.0592V}}{\text{n}}{\text{log}}_{\text{10}}\text{Q}\\ & & {\text{=E}}^{\text{o}}\left(K^{+}/K\right){\text{- 0.0592Vlog}}_{\text{10}}\frac{\text{1}}{\left(K^{+}\right)}\\ & & {\text{= - 2.925V - 0.0592Vlog}}_{\text{10}}\left[\frac{\text{1}}{\text{0.05}}\right]\\ & & \text{= - 2.295V -}\left(\text{0.0592V}\right)\left(\text{1.3010}\right)\\ & & \text{= - 2.295V - 0.077V}\\ & & \text{= - 3.00V}\end{array}$

This is more negative than reduction potential for ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$in pure water i.e., ${\text{E}}^{\text{0}}\text{= - 0.414V}$Hence, ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ion reduced.

Step-3: Reduction potential of third half-reaction

This is more positive than reduction potential for water i.e., $E^o=0.815V$$B{n}_2$ Hence, more easily oxidised than$O_2\left(g\right)$ which is generated at anode

Step-4: Final reaction

At anode:

$6H_{2}O\to O_2+4H_3{O}^{+}\left({10}^{-7}M\right)+4e^{-}$

At cathode:

$2H_3{O}^{+}\left({10}^{-7}M\right)+2e^{-}\to H_2+2H_{2}O$

Hence,

The products that form at the anode and the cathode are : Br₂ and K respectively.

At cathode :

${\text{K}}^{\text{+}}{\text{(aq) +e}}^{\text{-}}\to \text{K(s)}$

At anode :

$2B{r}^{-}\left(aq\right)+2e^{-}\to Br2\left(l\right)$

$$\begin{gathered} {\text{K}}^{\text{+}}{\text{(aq) +e}}^{\text{-}}\to \text{K(s)} \\ 2B{r}^{-}\left(aq\right)+2e^{-}\to Br2\left(l\right) \end{gathered}$$