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1989 WAEC Chemistry Theory (a) Distinguish between a conductor and an electrolyte (b)(i) State Faraday's first law of electrolysis...

Chemistry
WAEC 1989

(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.

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Explanation

(a) A conductor is a solid metal which allows electricity to pass through it while an electrolyte is a compound in the molten or in solution which allows electric current to pass through it.

(b)(i) Faraday's first law of electrolysis states that the amount of element deposited at the electrodes is directly proportional to the quantity of electricity applied in the electrolysis.

(ii) Different quantity of electricity at different time is allowed to pass through a fixed concentration of CuSO\(_4\) solution using copper electrodes. It could be observed that the higher the quantity of electricity, the greater the amount of copper element deposited at the cathode for example.

(c) Quantity of electricity = Current x tim

i.e 0.45 ampere x 1500 secs.

0.45 Coulomb x 15000 Sec.

0.45 \(\frac{coulomb}{Sec}\) x \(\frac{15000}{1}\) = 675 coulombs

If 675 coulombs give 0.222g of x\(^{2+}\)  i.e. 1 coulomb will give \(\frac{0.222}{675}\)g x\(^{2+}\)

2 For 2 x 965000 coulombs = \(\frac{0.222}{675}\) x 2 x \(\frac{965000}{1}\) 

= 63.5g

(d) Electrolysis is used to convert electrical energy to chemical energy. It can also be used for electroplating


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