1.

Consider the following cell reaction : 2Fe_(s) + O_(2(g)) + 4H_((aq.))^(+) rarr 2Fe_((aq.))^(2+) + 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»

`1.47 V`
`1.77 V`
`1.87 V`
`1.57 V`

SOLUTION :`2Fe_(s) + O_(2(g)) + 4H_((AQ.))^(+) rarr 2Fe_((aq.))^(2+) + 2H_(2)O_((L))`
`n = 4` (no. of moles of electron involved)
From Nernst's equation,
`E_(cell) = E_(cell)^(@) - (0.059)/(n)log Q`
`= 1.67 - (0.0591)/(4)log'((10^(-3))^(2))/(0.1 XX (10^(-3))^(4))`
`= 1.67 xx 0.106`
`= 1.57 V`


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