Consider the following cell reaction 2Fe(s)+O2(g)+4H+(aq)⟶ 2Fe2+(aq)+2H2O(l),E∘=1.67V At Fe2+=10−3M,pO2=0.1atm and pH=3, the cell potential at 25oC is

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1.47 V

1.77 V

1.87 V

1.57 V

answer is D.

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Detailed Solution

The half reactions are Fe(s)⟶Fe2+(aq)+2e−×2O2(g)+4H++4e−⟶2H2O2Fe(s)+O2(g)+4H+⟶2Fe2+(aq)+2H2O(l)E=1.67−0.0594log⁡10−3210−34(0.1)=1.57V

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