1.

The half cell reactionfor the corrosion are , 2H^(+) + (1)/(2) O_2 + 2e^(-) to H_2O , E^(0) = 1.23 V and Fe^(2+) + 2e^(-) to Fe(s) , E^(0) = -0.44 V . Find the DeltaG^(0) ( in kJ) for the overall reaction

Answer»

`-76 K `
`-322kJ`
`-161 KJ`
`-152 kJ`

Solution :`Fe to Fe^(+2) + 2e^(-) , E^(0) = 0.44`........ (2)
(1) `2H^(oplus) + (1)/(2) O_2 + 2e^(-) to H_2O , (+ 1.23 xx 2) ""(2) Fe^(+2) + 2e^(-) to Fe , (-0.44 xx 2)`
(3) `Fe + 2H^(oplus) + (1)/(2) O_2 to Fe^(+2) + H_2O , (E^(0) xx 2) ,(3) = (1)-(2)`
` 2E^(0) = (1.23 xx 2) - (-0.44 xx 2) , E^(0) = (2.46 + 0.88)/(2) = (3.34)/(2) , E^(0) = 1.67V`
`DeltaG = -2 xx 96500 xx 1.67 = -322.3 KJ`


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