Question

A voltaic cell is constructed that uses the following reaction and operates at 298 K.

Zn(s) + Ni²⁺(aq) Zn²⁺(aq) + Ni(s)

(a) What is the emf of this cell under standard conditions?
V
(b) What is the emf of this cell when [Ni²⁺] = 3.51 M and [Zn²⁺] = 0.191 M?
V
(c) What is the emf of the cell when [Ni²⁺] = 0.137 M and [Zn²⁺] = 0.878 M?
V

Answer 1

Solution:

(a)

(i) Ni²⁺(aq) + 2e^-$\rightarrow$ Ni (s) _________E^o = -0.277 V

(ii) Zn2+(aq) + 2e- $\rightarrow$ Zn (s) _________E^o = -0.763 V

E^o_cell = Ered (Ni2+/Ni) - Ered (Zn2+ /Zn)

E^o_cell = -0.277 V - (-0.763 V)

E^o_cell= 0.48 V

At standard condition emf is = 0.48V

(b) when [Ni²⁺] = 3.51 M and [Zn²⁺] = 0.191 M

E_cell= E^o_cell- (RT/nF) ln [Oxidised]/[Reduced]

E_cell= 0.48 - (8.314 x 298 / 2 x 96500) ln [0.191]/[3.51]

E_cell= 0.48 - (8.314 x 298 / 2 x 96500) ln [0.191]/[3.51]

E_cell= 0.48 - (0.0592 / 2) log [0.191]/[3.51]

E_cell= 0.48 - (0.0296) x (-1.26)

E_cell= 0.48 + 0.037

E_cell= 0.48 + 0.037

E_cell= 0.517V

E_cell= 0.52 V

emf is = 0.52 V

(c) when [Ni²⁺] = 0.137 M and [Zn²⁺] = 0.878 M

E_cell= E^o_cell- (RT/nF) ln [Oxidised]/[Reduced]

E_cell= 0.48 - (8.314 x 298 / 2 x 96500) ln [0.878]/[0.137]

E_cell= 0.48 - (0.0592 / 2) log [5.69]

E_cell= 0.48 - (0.0296) x (0.755)

E_cell= 0.48 - 0.02235

E_cell= 0.48 + 0.037

E_cell= 0.4577 V

E_cell= 0.46 V

emf is = 0.46 V