What is the ΔG∘ for the following reaction? Cu2+(aq)+2Ag(s)→Cu(s)+2Ag+(aq)
Given:E0Cu2+/Cu=0.34V,E0Ag/Ag+=−0.8V
A
-44.5 kJ
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B
44.5 kJ
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C
-89 kJ
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D
89 kJ
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Solution
The correct option is D 89 kJ Point to remember: ⇒Δ0G=−nFE0cell
Given informations in question : ⇒Reductionhalfcell Cu2+(aq)+2e−→Cu(s) ⇒Oxidationhalfcell Ag(s)→Ag+(aq)+e−
So from these half reactions we find that n=2
To calculate ΔG we need to find Eocell Eocell = Eocathode−Eoanode Eocell = EoCu2+/Cu−EoAg+/Ag Eocell=0.34−0.8=−0.46
On substituting Δ0G=−2×96500×(−0.46)=88780≈89kJ