Chemistry Electrochemistry

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New answer posted

2 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

The reaction is N? O? (g)? 2NO? (g).

Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = 2 - 1 = 1.

The relationship between Kp and Kc is Kp = Kc (RT)^Δn_g.

600.1 = 20.4 * (0.0831 * T)¹

T = 600.1 / (20.4 * 0.0831) = 353.99 K.

The answer, rounded off, is 354 K.

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

ΔG° = –nFE°
∴ 17.37 x 10³ = –3 x 96500 x E°
∴ E° = –6 x 10? ² V

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

For concentration cell E? cell = 0
Anode: Cu (s) → Cu²? (aq)?
Cathode: Cu²? (aq)? → Cu (s)
Overall: Cu²? (aq)? → Cu²? (aq)?
As ΔG = −nFEcell
If ΔG = -ve than Ecell is positive
Ecell = E? cell − (0.059/2) log (C? /C? )
Ecell = − (0.059/2) log (C? /C? )
Ecell > 0 ⇒ C? > C?

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Charge (q) = (it/96500) F = (1x15x60)/96500 = 900/96500 = 9/965 F = 0.0093F
No. of moles of Au? = 0.025 & No. of moles of Ag? = 0.025
Species with higher value of SRP will get deposited first at cathode.
(i) Au? (aqs) + e? → Au (s)
0.025 0.0093 mole
So only Au will get deposited

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

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

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

According to Faraday law

  W = Z I t * η Where η = efficiency

or W = E 96500 * 1 * t * η

Putting values

. 104 = ( 52 / 3 ) * 2 * 8 * 60 * η 96500 η = 0.6

So percentage efficiency = 60 %

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