Question

11. (12pts) For the redox reaction given below, Ni(s) + 2 H(aq) + Ni2+(aq) + H2(e) (show your work in detail) a) Write oxidation and reduction half cell reactions b) Write the electrochemical cell notation c) Predict whether this reaction will be spontaneous. Explain. d) CalculateGº 6 Page

Answer 1

Answer: a) Anode:
`Ni\rightarrow Ni^(2+)+2e^-`

Cathode :
`2H^++e^-\rightarrow H_2`

b)
`Ni/Ni^(2+)//H^+/H_2`

c) As
`E_(cell)=+0.25V` , the reaction is spontaneous.

d)
`\Delta G^0=-48250J`

Step-by-step explanation:

`Ni+2H^(+)\rightarrow Ni^(2+)+H_2`

a) Here Ni undergoes oxidation by loss of electrons, thus act as anode. Hydrogen undergoes reduction by gain of electrons and thus act as cathode.

Anode:
`Ni\rightarrow Ni^(2+)+2e^-`

Cathode :
`2H^++e^-\rightarrow H_2`

b) The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a slat bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

`Ni/Ni^(2+)//H^+/H_2`

c)
`E^0=E^0_(cathode)- E^0_(anode)`

Where both
`E^0` are standard reduction potentials.

`E^0_([Ni^(2+)/Ni])= -0.25V`

`E^0_([H^(+)/H_2])=+0.0V`

`E^0=E^0_([H^(+)/H_2])- E^0_([Ni^(2+)/Ni])`

`E^0=+0.0-(-0.25V)=0.25V`

`E_(cell)`= +ve, reaction is spontaneous

`E_(cell)`= -ve, reaction is non spontaneous

`E_(cell)`= 0, reaction is in equilibrium

Thus as
`E_(cell)=0.25V` , the reaction is spontaneous.

d) The standard emf of a cell is related to Gibbs free energy by following relation:

`\Delta G^0=-nFE^0`

`\Delta G^0` = standard gibbs free energy

n= no of electrons gained or lost

F= faraday's constant

`E^0` = standard emf

`\Delta G^0=-2* 96500* (0.25)=-48250J`

Thus value of Gibbs free energy is -48250 Joules.