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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2018
ChemistryHardnumerical

The surface of copper gets tarnished by the formation of copper oxide.N2gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, theN2gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:2Cus+ H2Og Cu2O(s) + H2(g)PH2Is the minimum partial pressure ofH2(in bar) needed to prevent the oxidation at 1250 K. The value oflnpH2is ____. (Given: total pressure = 1 bar, R (universal gas constant)= 8 J K-1 mol-1,ln10 = 2.3. Cu(s) and Cu2O(s) are mutually immiscible. At1250 K: 2Cus+12 O2g Cu2O(s); G = - 78,000 J mol-1H2g+12O2g H2O(g);      = - 1,78,000 J mol-1; G is the Gibbs energy)

Answer:
-14.60
Solution:

2Cus+14O2g1 Cu2OsG0=-78 kJ
H2g+12O2H2Og                        Go=-178 kJ×-1
Hence,2Cus+H2OgCu2O+H2g      Go=+100kJ
G=Go+RTlnQ
0=+100+81000×1250lnPH2PH2O
-100×10008=1250lnPH21100×1
lnPH2=-14.6

Stream:JEE_ADVSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells
2mℹ️ Source: PYQ_2018

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