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) $\Delta G^0=-48250J$

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

$\Delta G^0=-2\times 96500\times (0.25)=-48250J$

Thus value of Gibbs free energy is -48250 Joules.