Calculate the useful work of the reaction Ag(s)+1/2Cl2(g)→AgCl(s). Given E0cl2/Cl−=+1.36V,E0Ag|AgCl|Cl−=0.22V if PCl2=1atm and T=298K.
A
110 kJ/mol
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B
220 kJ/mol
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C
55 kJ/mol
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D
1000 kJ/mol
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Solution
The correct option is A 110 kJ/mol (1) AgCl(s)+e→Ag(s)+Cl−E0=0.22V (2) 1/2Cl2+e→Cl−E0=1.36V We get Ag(s)+1/2Cl2(g)→AgCl(s)E0cell=1.14V ∴ΔG=−nEF0=−(1)(96500)(1.14)=−110kJ/mol