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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 and pH=3, the cell potential at `25^(@)C` isA. `1.27V`B. `1.77V`C. `1.87V`D. `1.57V` |
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Answer» Correct Answer - D `[H^(+)]=10^(-3),Q=([Fe^(2+)])/([H^(+)]^(4)P_(O_(2))),n=4` `E=1.67-(0.06)/(4)log.((10^(-3))^(2))/((10^(-3))^(4)(10^(-1)))` `=1.67-(0.06)/(4)xx7` `=1.57V` |
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