(a) Distinguish between a conductor and an electrolyte (b)(i) State Faraday's first law of electrolysis (ii) Describe how you would investigate Faraday's la...

Assessment: WAEC SSCE - Chemistry - 1989 (Essay) Subject: Chemistry

Question 1 Report

(a) Distinguish between a conductor and an electrolyte

(b)(i) State Faraday's first law of electrolysis

(ii) Describe how you would investigate Faraday's law of electrolysis, using copper (II) tetraoxosulphate (VI) solution and copper electrodes.

(c) 0.222g of a divalent metal is deposited when a current of 0.45 ampere is passed through a solution of its salt for 25 minutes using appropriate electrodes. Calculate the relative atomic mass of the metal. 1F = 96500C mol\(^{-1}\)

`(d) State two applications of electrolysis.

Answer Details

(a) Conductor vs electrolyte

A conductor is a substance that allows electricity to pass through it without undergoing any chemical change; it conducts by the movement of free electrons (for example, a metal). An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it; it conducts by the movement of ions.

(b)(i) Faraday's first law

The mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.

(ii) Investigation

  • Weigh a clean, dry copper cathode and record its mass.
  • Set up an electrolytic cell containing copper(II) tetraoxosulphate(VI) solution with two copper electrodes, connected in series with an ammeter, a rheostat and a switch.
  • Pass a steady current I (read on the ammeter) for a measured time t (with a stopwatch); the charge is Q = It.
  • Remove, wash, dry and re-weigh the cathode to find the mass of copper deposited.
  • Repeat for different values of Q, then plot a graph of mass deposited against Q. A straight line through the origin confirms that mass is proportional to charge.

(c) Relative atomic mass

\[Q = It = 0.45 \times (25 \times 60) = 675\,C\]

For a divalent metal, M2+ + 2e- → M, so 2 × 96500 C deposit 1 mole of metal.

\[\text{moles of metal} = \frac{675}{2 \times 96500} = 3.497 \times 10^{-3}\,mol\]

\[\text{RAM} = \frac{0.222}{3.497 \times 10^{-3}} \approx 63.5\]

The relative atomic mass of the metal is about 63.5 (the metal is copper).

(d) Two applications of electrolysis

  • Electroplating of metals (for example, coating with silver, chromium or nickel).
  • Purification (refining) of metals such as copper; also the extraction of reactive metals such as aluminium and sodium.

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