Chemistry Electrochemistry
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New answer posted
2 months agoContributor-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 agoNew answer posted
2 months agoContributor-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 agoContributor-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 agoContributor-Level 9
Eext > 1.1 V, cell reaction is reversed
i.e. Cu → Cu²? anode
Zn²? + 2e? → Zn cathode
New answer posted
2 months agoContributor-Level 10
According to Faraday law
Where efficiency
or
Putting values
So percentage efficiency
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