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Commercial electrolysis is performed on both moltenand aqueoussolutions. Identify the anode product, cathode product, species reduced, and species oxidized for the anode.

(a) molten electrolysis and

(b) aqueous electrolysis.

Short Answer

Expert verified

(a) For molten electrolysis the anode product and the cathode product are Cl2and Narespectively.

(b) For aqueous electrolysis the anode product and the cathode product are Cl2and H2+ OH-respectively.

Step by step solution

01

Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

02

(a) Molten Electrolysis

In molten electrolysis of NaCl, sodium metal is formed at the cathode, as it is reduced from Na+to Nametal, while Chlorine gas is formed at the anode, as theCl- atoms are oxidized and form gas.

Due to overvoltage, not all ions can be converted into their free element form via electrolysis, including these:

  • Cations of metals from Group1A .
  • Cations of metals from Group2A and Aluminium from group 3A.
  • The fluoride ion cannot be oxidized, while the rest of the halides can be.
  • Common oxyanions such as SO42 -, CO32 -, NO3-, PO43 -cannot be oxidized because they are at their highest oxidation states as they are. Water is oxidized to O2,H+instead.

Therefore, at the anode, chlorine is the product, and at the cathode, sodium is the product.

03

(b) Aqueous Electrolysis

In the electrolysis of aqueousNaCl , metal will be formed at the cathode. This is because itNa+is a group 1Acation, water is reduced instead. However, at the anode, chlorine gas is formed from chloride ions.

Therefore, at the anode, chlorine is the product, and at the cathode, hydrogen and hydroxide ion is the products.

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Most popular questions from this chapter

Comparing the standard electrode potentials (Eo)of the Group 1A(1)metalsLi, Na, and Kwith the negative of their first ionization energies reveals a discrepancy:

Ionization process reversed:M+(g) +e-โ‡ŒM(g) ( - IE)

Electrode reaction:M+(aq) +e-โ‡ŒM(s) (Eo)

Note that the electrode potentials do not decrease smoothly down the group, as the ionization energies do. You might expect that if it is more difficult to remove an electron from an atom to form a gaseous ion (largerIE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smallerEo), yetLi+(aq)is more difficult to reduce thanNa+(aq). Applying Hessโ€™s law, use an approach similar to that for a Born-Haber cycle to break down the process occurring at the electrode into three steps and label the energy involved in each step. How can you account for the discrepancy?

How does over voltage in๏ฌ‚uence the products in the electrolysis of aqueous salts?

Consider the following general voltaic cell:

Identify the (a) anode, (b) cathode, (c) salt bridge, (d) electrode at which leave the cell, (e) electrode with a positive charge, and electrode that gains mass as the cell operates (assuming that a metal plates out).

Use Appendix D to calculate theKspof AgCl.

A silver button battery used in a watch contains 0.75 gof zinc and can run until 80%of the zinc is consumed.

(a) How many days can the battery run at a current 0.85of microamps (10- 6amps)?

(b) When the battery dies, 95%of the Ag2Ohas been consumed. How many grams ofwas used to make the battery?

(c) IfAg costs $ 13.00per troy ounce (31.10 g), what is the cost of theconsumed each day the watch runs?

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