Question

during electrolysis of aqueous solution containing 500 ml of NaBr, the cathode was obtained o, 56 l hydrogen under normal conditions. Find the mass in grams of the substance was obtained in the anode and the pH of the solution.

Answer 1

Question

During electrolysis of aqueous solution containing 500 ml of NaBr, the cathode was obtained o, 56 l hydrogen under normal conditions. Find the mass in grams of the substance was obtained in the anode and the pH of the solution.

Given

$V_{\text{solution}} = 500 \, \text{ml}$

$V_{H2} = 0.56 \, \text{l}$ (normal condition)

$m = ?$

$pH = ?$

Solution

NaBr dissociates in the solution as follows

$$\mathrm{NaBr} \leftrightarrow \mathrm{Na}^{+} + \mathrm{Br}^{-}$$

The electrode reactions are

$A(+): 2\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2} + 2\mathrm{e}^{-}$

$C(-): 2\mathrm{H}_{2}\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2} + 2\mathrm{OH}^{-}$

The substance obtained in the anode is bromine $(\mathrm{Br}_2)$

Overall reaction is

$$2 \mathrm{NaBr} + 2 \mathrm{H}_{2}\mathrm{O} \rightarrow 2 \mathrm{NaOH} + \mathrm{Br}_{2} + \mathrm{H}_{2}$$

Hydrogen and bromine are formed with equimolar ratio. $V_{\mathrm{m}} = 22.4 \, \mathrm{l/mol}$ (under normal conditions), $M_{\mathrm{Br2}} = 159.8 \, \mathrm{g/mol}$. So, we may write a proportion

22.4 l/mol $(\mathrm{H}_{2}) - 159.8 \, \mathrm{g/mol} \, (\mathrm{Br}_{2})$

0.56 l $(\mathrm{H}_{2}) - m \, \mathrm{g} \, (\mathrm{Br}_{2})$

Whence

$$m = \frac{159.8 \cdot 0.56}{22.4} = 3.995 \, \mathrm{g}$$

Since NaOH is formed, solution has basic pH that is calculated by the following equation

$$pH = 14 - pOH = 14 + \log [\mathrm{OH}^{-}]$$

Molarity of hydrixyl ions is

$$[\mathrm{OH}^{-}] = n(\mathrm{OH}^{-}) / V_{\text{solution}}$$

By the reaction equation we know that 1 mole of $\mathrm{H}_{2}$ corresponds to 2 moles of NaOH. So, we have the proportion

22.4 l/mol $(\mathrm{H}_{2}) - 2 \, \mathrm{moles} \, (\mathrm{NaOH})$

0.56 l $(\mathrm{H}_{2}) - n \, \mathrm{moles} \, (\mathrm{NaOH})$

Whence

$$n = \frac{0.56 \cdot 2}{22.4} = 0.05 \, \mathrm{moles}$$

$n(\mathrm{OH}^{-}) = n(\mathrm{NaOH}) = 0.05 \, \mathrm{moles}$

Thus, the solution molarity

$$[\mathrm{OH}^{-}] = \frac{0.05}{0.5} = 0.1 \, \mathrm{mol/l}$$ $$pH = 14 + \log 0.1 = 14 - 1 = 13$$

Answer: $m(\mathrm{Br}_{2}) = 3.995 \, \mathrm{g}, \, \mathrm{pH} = 13$