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The rusting of iron takes place as follo...

The rusting of iron takes place as follows :
`2H^(+) + 2e^(-) + (1)/(2)O_(2)(g) rarr H_(2)O (l) E^(@) = + 1.23 V`
`Fe^(2+) + 2e^(-) rarr Fe(s) E^(@) = - 0.44 V`
Calculate `Delta G^(@)` for the net process

A

`- 322 kJ mol^(-1)`

B

`- 161 kJ mol^(-1)`

C

`- 152 kJ mol^(-1)`

D

`- 76 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Fe(s) rarr Fe^(2+) + 2e^(-) Delta G_(1)^(@)`
`2H^(+) + 2e^(-) + (1)/(2) O_(2) rarr H_(2)O (l) Delta G_(2)^(@)`
`Fe (s) + 2H^(+) + (1)/(2) O_(2) rarr Fe^(2+) + H_(2)O Delta G_(3)^(@)`
`Delta G_(1)^(@) + Delta G_(2)^(@) = Delta G_(3)^(@)`
:. `Delta G_(3)^(@) = (- 2F xx 0.44) + (- 2F xx 1.23)`
`= (- 2 xx 96500 xx 0.44) + (- 2 xx 96500 xx 1.23)`
`= - 84920 - 237390`
`= - 322.310 kJ or - 322 kJ`
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