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The half cell reaction for rusting of ir...

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

A

`-76`

B

`-322`

C

`-161`

D

`-152`.

Text Solution

Verified by Experts

The correct Answer is:
B
`E^(@)=1.23+0.44=1.67V`
`DeltaG^(@)=-nFE^(@)`
`=02xx96500xx1.67=-322310J`
`=-322.31kJ`
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The following reaction occurs during rusting of iron 2H^(+)+2e+(1)/(2)O_(2)rarrH_(2)O,E^(@)=+1.23V Fe^(2+)+2erarr Fe(s), E^(@)=0.44V Calculate magnitude of DeltaG^(@)(kJ) for the net process Fe(s)+2H^(+)+(1)/(2)O_(2)rarrFe^(2+)+H_(2)O

The half-cell reaction for the corrosion, {:(2H^(+)+,(1)/(2)O_(2)+2e^(-) rarr H_(2)O,,,E_(@) = 123 V),(,Fe^(2+)+2e^(-) rarrFe(s),,,E^(@) = -0.44 V):} Find the Delya G^(@) (in kJ) for the overall reaction :

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))

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.

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