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
Among Zn and Cu, which would occur more readily in nature as metal and which as an ion?
E°cell for the given redox reaction is 2.71 V
Mg(s) + Cu2+ (0.01 M) → Mg2+ (0.001 M) + Cu(s)
Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V
For the electrochemical cell:
\[\ce{M | M+ || X- | X}\];
\[\ce{E^{\circ}_{{M^{+}/{M}}}}\] = 0.44 V,
\[\ce{E^{\circ}_{X/X^-}}\] = 0.33 V
Which of the following is TRUE for this data?
A solution of CuSO4 is electrolysed using a current of 1.5 amperes for 10 minutes. What mass of Cu is deposited at cathode? [Atomic mass of Cu = 63.7]
A copper electrode is dipped in 0.1 M copper sulphate solution at 25°C. Calculate the electrode potential of copper.
[Given: \[\ce{E^0_{{Cu^{2+}|Cu}}}\] = 0.34 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.
If 0.5 amp current is passed through acidified silver nitrate then in 100 minutes the mass of silver, deposite on cathode is (eq. wt. of silver nitrate + 108).
Explain why the anode is of negative polarity in a galvanic cell.
What are electrochemical reactions?
Explain the types of electrochemical cells.
