Electrochemistry

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New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

K e q = ( 1 2 * 1 0 1 5 ) 1 / 2 = 2 . 2 4 * 1 0 ? 8

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

meqofl2=meqofNa2S2O3=20*0.002*1

2 * mmol of l2 = 0.4

mmol of l2 = 0.2 mmol

mmol of Cu2+ = 0.2 * 2 * 10-3

[ C u 2 + ] = 0 . 4 * 1 0 3 1 0 * 1 0 3 = 0 . 0 4 = 4 * 1 0 2 M

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

Eext>1.1V , cell reaction is reversed
i.e. Cu? Cu2+ anode
Zn2++2e-? Zn cathode

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7 months ago

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P
Payal Gupta

Contributor-Level 10

C r 2 O 7 2 + 6 e + 1 4 H + 2 C r 3 + + 7 H 2 O  

Here, 6F electricity is required to reduce 1 mol  C r 2 O 7 2 to Cr3+.

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P
Payal Gupta

Contributor-Level 10

Using ; ksp = 108 S5

1.1 * 10-23 = 108 S5

S = ( 1 1 0 1 0 8 * 1 0 2 5 ) 1 / 5 M 1 0 5 M  

Specific conductance,

λ m 0 = κ 5 * 1 0 0 0 S m 2 m o l 1                              

= 3 * 10-3 Sm2 mol-1

So; x = 3

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Given cell constant lA=129m1

Ratio of ppm of two solution of KCl = Ratio of moles of per unit volume.

(ppm)KCl1 (ppm)KCl2=M1M2=74.5149.0=12

12=1000*103

 x = 1000

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Sol. Balmer series lies in the visible region.

New question posted

7 months ago

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New answer posted

7 months ago

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P
Payal Gupta

Contributor-Level 10

For the given cell, net redox reaction is as;

H 2 ( g ) + 2 A g + ( a q ) 2 H + ( a q ) + 2 A g ( s )               

n = 2

E c e l l 0 = + 0 . 5 3 3 2 V               

Using;  Δ G 0 = n F E c e l l 0  

= 2 * 9 6 4 8 7 * 0 . 5 3 3 2 J / m o l               

= 1 0 2 . 8 9 k J / m o l               

Now, for the reaction

1 2 H 2 ( g ) + A g + ( a q ) ? H + ( a q ) + A g ( s )               

n = 1

So; Δ G o = 1 * 9 6 4 8 7 * 0 . 5 3 3 2 J / m o l  

5 1 . 4 4 k J / m o l

New answer posted

7 months ago

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P
Pallavi Pathak

Contributor-Level 10

The NCERT Exemplar questions go beyond the textbook questions as they are designed to be more challenging and conceptually advanced. Practicing the NCERT exemplar provides a better understanding of complex topics like conductance and electrochemical cells. It also helps in better preparation for the entrance exams. The exemplar often involves multi-concept, application-based, and analytical problems.

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