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 reaction for the corrosion are , 2H^(+) + (1)/(2) O_2 + 2e^(-) to H_2O , E^(0) = 1.23 V and Fe^(2+) + 2e^(-) to Fe(s) , E^(0) = -0.44 V . Find the DeltaG^(0) ( in kJ) for the overall reaction

The half-cell reaction for the corrosion. 2H^(+)+1/2O_(2)rarr H_(2)Ol,E^(@)=1.23 V Fe ^(2+)+2e ^(−) →Fe (s);E^(@) =−0.44V Find the triangle 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^(-) rightarrow Fe , E^(@) = -0.44V Delta^(@) ((inKJ) for the reaction is : Fe+2H^(+) + (1)/(2) O_(2)rightarrow Fe^(+2)+ H_(2) O ((l))

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 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

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