Home
Class 12
CHEMISTRY
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

`-76`

B

`-322`

C

`-122`

D

`-176`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaG^(@) =- nFE^(@)`
`Fe^(+2) +2e^(Theta) harr Fe(n = 2)`
`E_(cell)^(@) = E_(H^(+)//H_(2)O)^(@) - E_(Fe^(2+)//Fe)^(@)`
`= 1.23 - (-0.44) = 1.67V`
`DeltaG^(@) =- 2 xx (96500 xx 1.67)/(1000) =- 322.3 kJ`
`DeltaG^(@) =- 322 kJ//mol`
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reduction of NO_(3)^(c-) ion in an aqueous solution, E^(c-) is +0.96V , the values of E^(c-) for some metal ions are given below : i.V^(2+)(aq)+2e^(-)rarr V, " "E^(c-)=-1.19V ii. Fe^(3+)(aq)+3e^(-) rarr Fe, " "E^(c-)=-0.04V iii. Au^(3+)(aq)+3e^(-) rarr Au, " "E^(c-)=+140V iv. Hg^(2+)(aq)+2e^(-) rarr Hg, " "E^(c-)=+0.86V The pair (s) of metals that is // are oxidized by NO_(3)^(c-) in aqueous solution is // are

The magnitude ( but not the sign ) of the standard reduction potentials of two metals X and Y are : Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V When the two half cells of X and Y are connected to construct a cell, eletrons flow from X to Y . When X is connected to a standard hydrogen electrode (SHE) ,electrons flow from X to SHE . If standard emf (E^(c-)) of a half cell Y^(2)|Y^(o+) is 0.15V , the standard emf of the half cell Y^(o+)|Y will be

Given, standard electrode potentials, {:(Fe^(3+)+3e^(-) rarr Fe,,,E^(@) = -0.036 " volt"),(Fe^(2+)+2e^(-)rarr Fe,,,E^(@) = - 0.440 " volt"):} The standard electrode potential E^(@) for Fe^(3+) + e^(-) rarr Fe^(2+) is :

E^(c-) of some elements are given as : {:(I_(2)+2e^(-)rarr 2I^(c-),,,,E^(c-)=0.54V),(MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(c-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(c-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(c-)=0.1V):} a. Select the stronges reductant and weakes oxidant among these elements. b. Select the weakest reductant and strongest oxidant among these elements.

Given that, Co^(3+) +e^(-)rarr Co^(2+) E^(@) = +1.82 V 2H_(2)O rarr O_(2) +4H^(+) +4e^(-), E^(@) =- 1.23 V . Explain why Co^(3+) is not stable in aqueous solutions.

Cu^(+) + e rarr Cu, E^(@) = X_(1) volt, Cu^(2+) + 2e rarr Cu, E^(@) = X_(2) X_(2) volt For Cu^(2+) + e rarr Cu^(+), E^(@) will be :