NCERT solution Class 12 electrochemistry

NCERT Solutions

Class 12 Chemistry

Electrochemistry

NCERT Questions

Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
Q.15
Q.16
Q.17
Q.18

Q.5

Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s) |Mg²⁺(0.001M)||Cu²⁺(0.0001 M)|Cu(s).

(ii) Fe(s)|Fe²⁺(0.001M)||H⁺(1M)|H₂(g)(1bar)| Pt(s)

(iii) Sn(s)|Sn²⁺(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s)

(iv) Pt(s)|Br²(l)|Br^–(0.010 M)||H⁺(0.030 M)| H₂(g) (1 bar)|Pt(s).

(i) For the given reaction, the Nernst equation can be given as:

E_cell = E _cell^(-) – (0.0591/n) log [Mg²⁺]/ [Cu²⁺]

= {0.34 – (-2.36)} – (0.0591/2) log (.001/.0001)

=2.7 – (0.0591/2) log10

= 2.7 − 0.02955

= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

E_cell = E _cell^(-) – (0.0591/n) log [Fe²⁺]/ [H⁺]²

= {0-(0.44)} – (0.0591/2) log [0.001]/ [1²]

= (0.44 – 0.02955) (-3)

=0.52865 V

=0.53 V (approx.)

(iii) For the given reaction, the Nernst equation can be given as:

E_cell = E _cell^(-) – (0.0591/n) log [Sn²⁺]/ [H⁺]²

= {0-(0.14)} – (0.0591/2) log (0.050)/ (0.020) ²

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:

E_cell = E _cell^(-) – (0.0591/n) log [1]/ [Br^-]²/ [H⁺]²

= (0- 1.09) – (0.0591/2) log (1)/ [(0.010) ²(0.030)]

=-1.09 -0.02955 x log (1/0.0000009)

=-1.09 – 0.02955 x log (1/9 x 10-8)

=-1.09 -0.02955(0.0453 +7)

=-1.09 – 0.208

=-1.298V

Video solution:

Electrochemistry Class 12 Chemistry NCERT Chapter 2 NCERT Solutions

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NCERT solutions of Electrochemistry Class 12 Chemistry - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 12 Chemistry Electrochemistry NCERT solutions.