(a)(i)
Conductors | Electrolytes |
Elements, usually metalsNot decomposed by passage of electricity Electrons carry the current | Are compounds (acids, bases, salts)Decomposed by electric current Ions carry the electric current (any two) |
(ii) Application of electrolysis
- For purification of elements e.g. copper, gold, silver
- For protection of metals from corrosion / electroplating
- For manufacture of chemicals e.g. NaOH, NaCIO\(_{3}\)
- For extraction of metals e.g Al, Ca, K, Na, Mg, etc.
(iii) Electrolysis of dilute NaCl(\(^{aq}\)) using carbon electrodes.
At the Cathode H\(^{+}\) + e\(^-\) —> H
H + H —> H\(^2\) Or 2H\(^{+}\) + 2e\(^-\) —> H\(_2\)
At the anode OH\(^-\) —> OH + e\(^-\)
or 4OH\(^- \) ---> 2H\(^2\)O + O\(_2\) Or 4OH\(^-\) —> 2H\(_2\)O + O\(_2\) + 4e\(^{-}\).
(b)(i) An electrochemical cell is a device for producing an electric current by chemical reaction.
(ii) Examples of primary cell: Daniell cell, Leclanche cell, cadmin / weston mercury cell.
(iii) Mg + Fe\(^{2+}\) --> Mg\(^{2+}\) + Fe ;
Half cell reactions: Mg --> Mg\(^{2+}\) + 2e\(^{-}\)
Fe\(^{2+}\) + 2e\(^-\) --> Fe or Mg - 2e\(^-\) --> Mg\(^{2+}\)
(c)(i) To calculate the quantity of electricity produced by a current of 0.72 A in 3 hours 20 mins.
Q = It
t = (3 x 60 + 20) min x 60 sec = 12000 Sec.
Q = (0.72 x 12000) C = 8640 C.
(ii) If 1 dm\(^3\) of gas was evolved at the cathode during electrolysis of acidified water, volume of gas evolved at the anode will be 0.5 dm\(^3\) or 500cm\(^3\).
(d)(i) Calculation of excess CuO if 20g was warmed with 0.05 mole H\(_2\)SO\(_4\). CuO + H\(_2\)SO\(_4\) --> CuSO\(_4\) + H\(_2\)O. From the equation 0.05 mole of H\(_2\)SO\(_4\) reacts with 0.05 mole of CuO.
Molar mass of CuO = 64 + 16 = 80g.
Hence 0.05 mole CuO = (80 x 0.05)g = 4.0g.
Mass of CuO in excess = 20 - 4 = 16g.
(ii) Type of reaction involved in the reaction in (d)(i) is neutralisation / acid - base reaction.