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The half cell reactions for rusting of i...

The half cell reactions for rusting of iron are :
`2H^(+) + 2e^(-) + (1)/(2)O_(2) rightarrow H_(2)O (l)` ,` E^(@) = +1.23V`
`Fe^(2+) + 2e^(-) rightarrow Fe` , `E^(@) = -0.44V`
` Delta^(@)` ((inKJ) for the reaction is : `Fe+2H^(+) + (1)/(2) O_(2)rightarrow Fe^(+2)+ H_(2) O ((l))`

A

`-76kJ`

B

`-322kJ`

C

`-161J`

D

`-152kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
`Fe(s) to Fe^(2+)+2e^(-) " "triangleG^(1)^(@)`
`2H^(+)+2e^(-)+1/2O_(2) to H_(2)O(l), " "triangleG_(2)^(0)`
`=Fe(s)+2H^(+)+1/2O_(2) to Fe^(2+) +H_(2)O" "triangleG_(3)^(@)`
Applying `triangleG_(1)^(@)+triangleG_(2)^@=triangleG_(3)^@`
`triangleG_(3)^@=(-2F xx 0.44)+(-2F xx 1.23)`
`=(-2 xx 96500xx 0.44)+(-2 xx 96500xx 1.23)`
=322310J=-322 kJ
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