Advertisements
Advertisements
Question
Describe the electrolysis of molten NaCl using inert electrodes.
Advertisements
Solution
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.
2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.
3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pulls the electrons from the anode.
Cell reactions:
Na+ ions are attracted towards the cathode, where they combine with the electrons and are reduced to liquid sodium.
Cathode (reduction)
\[\ce{Na^+_{ (l)} + e^- -> Na_{(l)}}\]
E0 = – 2.71 V
Similarly, Cl– ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas.
Anode (oxidation)
\[\ce{2Cl^-_{ (l)} -> Cl2_{(g)} + 2e^-}\]
E0 = – 1.36 V
The overall reaction is,
\[\ce{2Na^+_{ (l)} + 2Cl^-_{ (l)} -> 2Na_{(l)} + Cl2_{(g)}}\]
E0 = 4.07 V

Electrolysis of molten NaCl
The negative E0 value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07 V to cause the electrolysis of molten NaCl.
In an electrolytic cell, oxidation occurs at the anode and reduction occurs at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is −ve and the anode is +ve.
APPEARS IN
RELATED QUESTIONS
How many faradays of electricity are required for the following reaction to occur
\[\ce{MnO^-_4 -> Mn^2+}\]
Is it possible to store copper sulphate in an iron vessel for a long time?
Given: \[\ce{E^0_{{Cu^{2+}|{Cu}}}}\] = 0.34 V and \[\ce{E^0_{{Fe^{2+}|{Fe}}}}\] = −0.44 V
Use the data given in below find out the most stable oxidised species.
`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V `E^0 (Cl_2)/(Cl^-)` = 1.36 V
`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V `E^0 (Cr^(3+))/(Cr)` = – 0.74 V
`E_(cell)^Θ` for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \[\ce{H^{+} (aq) + e^{-} -> 1/2 H_2 (g); E^Θ_{cell} = 0.00V}\]
(b) \[\ce{2H2O (1) -> O2 (g) + 4H^{+} (aq) + 4e^{-}; E^Θ_{cell} = 1.23V}\]
(c) \[\ce{2SO^{2-}_{4} (aq) -> S2O^{2-}_{8} (aq) + 2e^{-}; E^Θ_{cell} = 1.96V}\]
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, \[\ce{SO4^{2-}}\] ion will be oxidised to tetrathionate ion at anode.
`E_(cell)^Θ` = 1.1V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?
(i) 1.1 = `K_c`
(ii) `(2.303RT)/(2F) logK_c` = 1.1
(iii) `log K_c = 2.2/0.059`
(iv) `log K_c` = 1.1
Match the terms given in Column I with the units given in Column II.
| Column I | Column II |
| (i) Λm | (a) S cm-¹ |
| (ii) ECell | (b) m-¹ |
| (iii) K | (c) S cm2 mol-¹ |
| (iv) G* | (d) V |
Match the terms given in Column I with the items given in Column II.
| Column I | Column II |
| (i) Λm | (a) intensive property |
| (ii) ECell | (b) depends on number of ions/volume |
| (iii) K | (c) extensive property |
| (iv) ∆rGCell | (d) increases with dilution |
The electrochemical cell stops working after some time because
The correct order of the mobility of the alkali metal ions. In aqueous solultion is
Explain the types of electrochemical cells.
