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Zorluk: OrtaIndustrial Applications of Electrolysis

In an industrial chlor-alkali membrane cell, concentrated sodium chloride solution (brine) is electrolyzed using a constant current of 19.3 A19.3\text{ A} for 50 minutes50\text{ minutes}. What is the mass, in grams, of sodium hydroxide (NaOH\text{NaOH}) produced in the solution? [Molar mass of NaOH=40.0 g mol−1\text{NaOH} = 40.0\text{ g mol}^{-1}, Faraday constant F=96,500 C mol−1F = 96,500\text{ C mol}^{-1}]

Cevap: 24 g

Cevap

The mass of sodium hydroxide produced is 24.0 g24.0\text{ g}.
Converting the electrolysis time to seconds (3000 s3000\text{ s}) yields a total charge of Q=19.3 A×3000 s=57,900 CQ = 19.3\text{ A} \times 3000\text{ s} = 57,900\text{ C}. Dividing by Faraday's constant (96,500 C mol−196,500\text{ C mol}^{-1}) gives 0.6 mol0.6\text{ mol} of electrons. In the chlor-alkali process, reduction of water produces 1 mol1\text{ mol} of OH−\text{OH}^- ions per mole of electrons, producing 0.6 mol0.6\text{ mol} of NaOH\text{NaOH}. Multiplying by the molar mass (40.0 g mol−140.0\text{ g mol}^{-1}) gives a mass of 24.0 g24.0\text{ g}.

Adım Adım Çözüm

1
Convert electrolysis time into seconds and calculate total charge.
Q=19.3 A×(50×60 s)=57,900 CQ = 19.3\text{ A} \times (50 \times 60\text{ s}) = 57,900\text{ C}.
Faraday's equations require time in seconds (Q=I×tQ = I \times t).
2
Calculate the moles of electrons transferred using the Faraday constant.
n(e−)=57,900 C96,500 C mol−1=0.6 moln(e^-) = \frac{57,900\text{ C}}{96,500\text{ C mol}^{-1}} = 0.6\text{ mol}.
One Faraday (96,500 C96,500\text{ C}) corresponds to one mole of electrons.
3
Determine the stoichiometry of the cathode reaction.
Cathode reaction: 2H2O(l)+2e−→H2(g)+2OH(aq)−2\text{H}_2\text{O}_{(l)} + 2e^- \rightarrow \text{H}_{2(g)} + 2\text{OH}^-_{(aq)}. Thus, 1 mol e−1\text{ mol } e^- forms 1 mol OH−1\text{ mol } \text{OH}^-, giving 0.6 mol0.6\text{ mol} of NaOH\text{NaOH}.
During brine electrolysis, water is preferentially reduced at the cathode, generating hydroxide ions that pair with sodium ions.
4
Calculate the mass of sodium hydroxide produced.
Mass=0.6 mol×40.0 g mol−1=24.0 g\text{Mass} = 0.6\text{ mol} \times 40.0\text{ g mol}^{-1} = 24.0\text{ g}.
Mass is obtained by multiplying the number of moles by the molar mass.

Anahtar Kavram

Industrial Electrolysis Stoichiometry (Chlor-Alkali Process) and Faraday's First Law
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Industrial Applications of Electrolysis

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