(a)(i) Give two differences between a conductor and an electrolyte.
(ii) State three applications of electrolysis.
(iii) Write equation for the reaction at each electrode when a dilute solution of sodium chloride is electrolysed using carbon electrodes.
(b)(i) What is an electrochemical cell?
(ii) Give two examples of primary cells.
(iii) Split the following equation into two balanced hall cell equations. Mte + Fe(^{2+} to Mg^{2+} + Fe).
(c)(i) A current of 0.72 amperes was passed through dilute tetraoxosulphate (VI) acid for 3 hours 20 minutes. Calculate the quantity of electricity that was passed
(ii) If 1 dm(^3) of gas evolved at the cathode during the electrolysis of acidified water, what was the volume of gas evolved at the anode?
(d)(i) 20g of copper(II) oxide was warmed with 0.05 mole of tetraoxosulphate (VI) acid. Calculate the mass of copper (II) oxide that was in excess. The equation for the reaction: CuO(_{(s)}) + H(_2)SO(_{4(aq)}) –> CuSO(_{4(aq)}) + H(_2)O(_l) [0 = 16 ; Cu = 64]
(ii) What type of reaction was involved in (d)(i)?
Explanation
(a)(i)
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Conductors |
Electrolytes |
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Elements, usually metals Not 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)
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(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.