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The rusting of iron takes place as fol...

The rusting of iron takes place as follows
`2H^(opluse)+2e^(-)+1/2 O_(2)rarrH_(2)O(l),E^(-)=+1.23 V`
`Fe^(2+)(aq)+2e^(-)+2e^(-)rarrFe(S),E^(-1)=-0.44V triangleG^(-)` for the net process is

A

`-322 kJ mol^(-1)`

B

`-152 kJ mol^(-1)`

C

`-76 kJ mol^(-1)`

D

`-161 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
At cathode `2H^(opluse)(aq)+2e^(-)+1/2 O_(2)(g)rarrH_(2)O(l)`
At anode `Fe(s) rarr Fe^(2+) rarr Fe^(+) (aq) +2e^(-)`
`triangle G_(2)^(-) =- 2xxFe^(-) =(-) 2xxF(0.44)`
`2H^(+opluse)(aq)+1/2 O_(2) +Fe(s)rarrH_(2)O(l)+Fe^(2+) (aq)`
`triangle G_("net")^(-) =[-2F(1.123 )] +(-) 2F(0.44) ] =- 322.3 kJ "mol"^(-)`
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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.

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 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

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 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 :

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