(a)(i) Give two differences between a conductor and an electrolyte. (ii) State three applications of electrolysis. (iii) Write equation for the reaction at ...
Question 1 Report
(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)?
Answer Details
(a)(i) Differences between a conductor and an electrolyte
| Conductor (metallic) | Electrolyte |
|---|---|
| Conducts by flow of electrons | Conducts by movement of ions |
| Undergoes no chemical change while conducting | Is chemically decomposed while conducting |
| Conducts in the solid state | Conducts only when molten or in aqueous solution |
(a)(ii) Three applications of electrolysis
- Electroplating of metals.
- Purification (refining) of metals such as copper.
- Extraction of reactive metals (e.g. sodium, aluminium) and manufacture of chemicals (NaOH, Cl2).
(a)(iii) Electrode reactions for dilute NaCl (carbon electrodes)
Cathode (reduction): \( 2\text{H}^+_{(aq)} + 2e^- \to \text{H}_{2(g)} \)
Anode (oxidation): \( 4\text{OH}^-_{(aq)} \to \text{O}_{2(g)} + 2\text{H}_2\text{O}_{(l)} + 4e^- \)
(Because the solution is dilute, oxygen, not chlorine, is discharged at the anode.)
(b)(i) An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy (or, in electrolysis, electrical energy into chemical energy).
(b)(ii) Primary cells: the Daniell cell and the dry Leclanche (zinc-carbon) cell.
(b)(iii) Half-cell equations for Mg + Fe2+ \(\to\) Mg2+ + Fe
Oxidation: \( \text{Mg} \to \text{Mg}^{2+} + 2e^- \)
Reduction: \( \text{Fe}^{2+} + 2e^- \to \text{Fe} \)
(c)(i) Quantity of electricity
\( t = 3\,\text{h}\,20\,\text{min} = 12000\ \text{s} \)
\[ Q = It = 0.72 \times 12000 = 8640\ \text{C} \]
(c)(ii) At the cathode H2 is evolved, at the anode O2; by the equation \(2\text{H}_2\text{O} \to 2\text{H}_2 + \text{O}_2\) the volume ratio H2:O2 is 2:1. So for 1 dm3 of H2 at the cathode, the anode gives 0.5 dm3 of O2.
(d)(i) Mass of CuO in excess
CuO + H2SO4 \(\to\) CuSO4 + H2O (1 : 1)
\[ n(\text{CuO}) = \frac{20}{64+16} = \frac{20}{80} = 0.25\ \text{mol} \]
Only 0.05 mol reacts (limited by acid), so excess CuO = \(0.25 - 0.05 = 0.20\) mol.
\[ \text{mass in excess} = 0.20 \times 80 = 16\ \text{g} \]
(d)(ii) A neutralization (acid-base) reaction.
Download The App On Google Playstore
Everything you need to excel in your exams
