Discussion Forum

1)

Consider the following cell reaction.

  $2Fe (s)+O_{2}(g)+4H^{+}(aq) \rightarrow$

                   $2Fe^{2+}(aq)+2H_{2}O(l),$  E°=1.67 V

At       $[FE^{2+}]=10^{-3}$M, $P(O_{2})=0.1 atm$

and  pH=3 , the cell potential at 25° C is 

Answer:

Option D

Explanation:

The half reactions are

 $Fe(s) \rightarrow  Fe^{2+}(aq)+2e^{-} \times 2$

 $O_{2}(g)+4H^{+}+4e^{-} \rightarrow 2H_{2}O+2Fe(s)$

  $+O_{2}(g)+4H^{+} \rightarrow   2Fe^{2+}(aq)+2H_{2}O(l)$

  $E= E^{0}-\frac{0.059}{4} log \frac{(10^{-3})^{2}}{(10^{-3})^{4})(0.1)}=1.57V$