Question

( mathrm{pH}=-log left[mathrm{H}^{+}right] )
( 3=-log left[mathrm{H}^{+}right] )
-antilog ( 3=left[mathrm{H}^{+}right] ) Therefore, ( left[mathrm{H}^{+}right]=10^{-3} mathrm{M} )
Oxidation half reaction ( left.mathrm{Fe}(mathrm{s}) rightarrow mathrm{Fe}^{2+}+2 mathrm{e}^{-} quadright] times 2 )
( 4 mathrm{H}^{+}+mathrm{O}_{2}+4 mathrm{e}^{-} rightarrow 2 mathrm{H}_{2} mathrm{O} )
Number of electrons involved, ( n=4 )
Cell potential is given as,
( mathrm{Ecell}=mathrm{E}^{circ} mathrm{cell}-frac{0.529}{mathrm{n}} log frac{left[mathrm{Fe}^{2+}right]^{2}}{left[mathrm{H}^{+}right]^{4}} )
( mathrm{Ecell}=1.67-frac{0.529}{4} log frac{left[10^{-3}right]^{2}}{left[10^{-3}right]^{4}} )
Ecell ( =1.67-0.13225 log 10^{6} )
( mathrm{Ecell}=1.67-0.13225 times 6 )
( mathrm{Ecell}=1.67-0.7935=0.8765 mathrm{V} )

# islaht of the reaction is Consider the following cell reaction 2Fe(s)+02(g) + 4H+ (aq) →2Fe2+ (aq) + 2H2O(1),E° = 167V At [Fe2+]=10-M,Po, =0.1 atm and pH = 3, the cell potential at 25°C is The standard emf. of a colla

Solution