Consider the following cell reation `:`
`2Fe(s)+O_(2)(g)+4H^(o+)(aq) rarr 2Fe^(2+)(aq)+2H_(2)O(l)" "E^(c-)=1.67V`
`At[Fe^(2+)]=10^(-3)M,p(O_(2))=0.1atm` and `pH=3` .
The cell potential at `25^(@)C` is

Options

A. 1.47 V
B. 1.77 V
C. 1.87 V
D. (c) 1.87 V (d)

Answer

D

Explanation

Here n = 4 and ` [ H^ + ] = 10 ^ ( - 3 ) ` (as pH = 3 )
Applying Nernst equation
` E = E^@ - ( 0.059 ) /( n ) log ([Fe^ ( 2 + ) ] ^ 2 )/([ H^+ ] ^ 4 (p _ (o _ 2 )) ) `
` = 1.67 - (0.059)/( 4 ) log (( 10 ^ ( - 3 ) ) ^ 2 ) /( ( 10 ^ ( -3) ) ^ 4 l xx 0.1 ) `
` = 1.67 - (0.059) /( 4 ) log 10 ^ 7 `
` = 1.67 - 0.103 = 1.567`