Q.

Consider the following cell reaction 
2Fe(s)+O2(g)+4H+(aq) 2Fe2+(aq)+2H2O(l),E=1.67V
 At Fe2+=103M,pO2=0.1atm and pH=3, the cell potential at 25oC is

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a

1.57 V

b

1.47 V 

c

1.77 V

d

1.87 V

answer is D.

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Detailed Solution

The half reactions are Fe(s)Fe2+(aq)+2e×2
O2(g)+4H++4e2H2O2Fe(s)+O2(g)+4H+2Fe2+(aq)+2H2O(l)
E=1.670.0594log10321034(0.1)=1.57V

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