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Consider the following cell reaction `2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq),2H_(2)O(l),E^(@)=1.67V` `At|Fe^(2+)| =10^(-3) M,P(O_(2))=0.1` atm pH =3, the cell potential at `25^(@)C` isA. 1.47VB. 1.77VC. 1.87VD. 1.57V |
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Answer» Correct Answer - D Applying Nernst equation of the given reaction `E_("cell")=E_("cell")^(@)-(0.0591)/(n)"log"([Fe ^(+2)]^ (2))/(P_(O_(2))xx[H^(+)]^(4))` For the given reaction, n=4 pH=3 means `[H^(+)]=10^(-3)M` `therefore E_("cell")=1.67-(0.059)/(4)"log"((10^(-3))^(2))/(0.1xx(10^(-3))^(4))` `=1.67-(0.059)/(4)"log"((10^(-1))^(2))/(0.1xx(10^(-3))^(4))` `=1.69-(0.059)/(4)"log"10^(7)` `=1.67-0.10=1.57V` |
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