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,
Ecell ( =mathrm{E}^{circ} ) 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} )

# Consider the following cell reaction 2Fe(s) +02(g) + 4H*(aq) →2Fe2+ (aq) + 2H,0(1), E = 167V At [Fe2+1=10-3M,Po2 = 0.1 atm and pH = 3, the cell potential at 25°C is

Solution