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The rusting of iron takes place as follo...

The rusting of iron takes place as follows `:`
`2H^(o+)+2e^(-) +(1)/(2)O_(2)rarr H_(2)O(l)," "E^(c-)=+1.23V`
`Fe^(2+)+2e^(-) rarr Fe(s)," "E^(c-)=-0.44V`
Calculae `DeltaG^(c-)` for the net process.

A

`-322kJ mol^(-1)`

B

`-161 kJ mol^(-1)`

C

`-152kJ mol^(-1)`

D

`-76 kJ mol^(-1)`

Text Solution

Verified by Experts

(D)Anode : `2OH^(-) rarr H_2O+1//2O_2+2e`,
Cathode : `H^(+)+2e rarr H_2`
Eq. of `O_2` formed = `(9.65xx10xx60xx60)/96500` =0.36
`:.` Mole of `O_2` formed =`0.36/4`= 0.09,
`C_8H_18+(25)/2 O_2 rarr 8CO_2+9H_2O`
`:.` Mole of `C_8H_(18)` = `25/2xx0.09`= `7.2 xx10^(-3)`
`:.V_(C_8H_(18))` = `7.2 xx10^-3xx22400` =161.2 mL
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The half cell reaction for rusting of iron are: 2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V DeltaG^(@) (in KJ) for the reaction is

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^(-) + (1)/(2)rightarrow H_(2)O_(2) ((l)) , E^(@) = -0.44V Delta^(@) ((inKJ) for the reaction is : Fe+2H^(+) + (1)/(2) O_(2)rightarrow Fe^(+2)+ H_(2) O ((l))

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The rusting of iron takes place as 2H^++2e+1/2O_2rarrH_2O(l),E^@=+1.23V Fe^(2+)+2e rarrFe(s),E^@=-0.44V Thus, DeltaG^@ for the net process is

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