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Electrochemistry · Question 64

AIPMT 2002MCQMedium📈 Topic in 0% papers+4−1
1

In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam because

  1. Hg is more inert than Pt
  2. more voltage is required to reduce H+ at Hg than at Pt
  3. Na is dissolved in Hg while it does not dissolve in Pt
  4. concentration of H+ ions is larger when Pt electrode is taken

Answer

(2) more voltage is required to reduce H+ at Hg than at Pt

Solution

At the cathode, Na+ and H+ (from water) compete for reduction.

At a Pt cathode, H+ (water) is reduced more easily ⇒ H2 is liberated.

As per NCERT, the discharge of hydrogen at a mercury cathode needs a high overpotential, so at a Hg cathode Na+ is reduced instead and the sodium dissolves in mercury to form sodium amalgam.

Na+ + e- + Hg Na-amalgam

∴ More voltage is required to reduce H+ at Hg than at Pt.

Answer: (2)

AIPMT 2002

Keep practising — Electrochemistry

2

Standard electrode potentials are Fe2+/Fe [ E = -0.44 V] and Fe3+/Fe2+ [ E = 0.77 V]; If Fe2+, Fe3+ and Fe blocks are kept together, then

  1. Fe3+ increases
  2. Fe3+ decreases
  3. Fe2+/Fe3+ remains unchanged
  4. Fe2+ decreases.
AIPMT 2001
3

The most convenient method to protect the bottom of ship made of iron is

  1. coating it with red lead oxide
  2. white tin plating
  3. connecting it with Mg block
  4. connecting it with Pb block
AIPMT 2001
4

Cell reaction is spontaneous when

  1. is negative
  2. is positive
  3. E red is positive
  4. E red is negative
AIPMT 2000
5

Ionic conductances of Ba2+ and Cl- ions at infinite dilution are 127 and 76 ohm-1 cm2 mol-1 respectively. Equivalent conductance of BaCl2 at infinite dilution is

  1. 139.5
  2. 101.5
  3. 203
  4. 279
AIPMT 2000
6

For the disproportionation of copper 2Cu+ Cu2+ + Cu, E is (Given E for Cu2+/Cu is 0.34 V and E for Cu2+/Cu+ is 0.15 V.)

  1. 0.49 V
  2. -0.19 V
  3. 0.38 V
  4. -0.38 V
AIPMT 2000

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