An electrochemical cell consists of two metallic electrodes (anode and cathode) dipping in electrolytic solutions. Oxidation takes place at the anode and reduction takes place at the cathode. These cells are of two types:
1) Galvanic cells, in which the chemical energy of a spontaneous redox reaction is converted into electrical energy. Eocell = Eocathode - Eoanode, where Eocell is the standard potential of the cell, and ΔrGo = -nFEocell, where ΔrGo is the standard Gibbs energy change of the cell reaction.
2) Electrolytic cells, in which electrical energy is used to carry out a non-spontaneous redox reaction. The amount of substance produced at an electrode depends on the quantity of electricity passed, Q = I × t (where I is the current in ampere and t is the time in seconds). One faraday is the charge carried by 1 mole of electrons, that is 96500 C.
The conductivity (κ) of an electrolytic solution depends on the nature and the concentration of the electrolyte. According to Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, for example Λom(NaCl) = λoNa+ + λoCl–, where λo is the limiting molar conductivity of an ion. The law can be used to calculate the limiting molar conductivity of weak electrolytes.
In electrolysis of an aqueous solution of sodium sulphate, 2.4 L of oxygen at STP was liberated at anode. The volume of hydrogen at STP liberated at cathode would be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.8 L
In the electrolysis of aqueous \( \ce{Na2SO4} \), neither \( \ce{Na^{+}} \) nor \( \ce{SO4^{2-}} \) is discharged; water is reduced at the cathode and oxidised at the anode:
\( \displaystyle \text{Cathode: } \ce{4H2O(l) + 4e- -> 2H2(g) + 4OH-(aq)} \)
\( \displaystyle \text{Anode: } \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-} \)
\( \displaystyle \text{Net: } \ce{2H2O(l) -> 2H2(g) + O2(g)} \)
The same charge (4 mol of electrons) gives 2 mol of \( \ce{H2} \) for every 1 mol of \( \ce{O2} \). At the same temperature and pressure, volumes are in the ratio of moles:
\( \displaystyle V_{\ce{H2}} = 2 \times V_{\ce{O2}} = 2 \times 2.4\ \mathrm{L} = 4.8\ \mathrm{L} \)
Shortcut: Electrolysing water always gives hydrogen and oxygen in the volume ratio 2 : 1.