Question

In a galvanic cell, one half-cell consists of a lead strip dipped into a 1.00 M solution of Pb(NO₃)₂. In the second half-cell, solid neodymium is in contact with a 1.00 M solution of Nd(NO₃)₃. Pb is observed to plate out as the galvanic cell operates, and the initial cell voltage is measured to be 2.197 V at 25°C.


(a) Write balanced equations for the half-reactions at the anode and the cathode. Show electrons as e^-. Use the smallest integer coefficients possible and the pull-down boxes to indicate states. If a box is not needed, leave it blank.

Half-reaction at anode (do not multiply by factor):

---------- (aq) (s) (l) (g)

+ --------

(aq) (s) (l) (g)

= --------

(aq) (s) (l) (g)

+ -------

(aq) (s) (l) (g)

Half-reaction at cathode (do not multiply by factor):

-------- (aq) (s) (l) (g)

+ ---------

(aq) (s) (l) (g)

=

--------- (aq) (s) (l) (g)

+ ----------

(aq) (s) (l) (g)

(b) Calculate the standard reduction potential of a Nd³⁺|Nd half-cell. The standard reduction potential of the Pb²⁺|Pb electrode is -0.126 V.

---------- V

Answer 1

First, let us defin the ions in solution

Pb(NJO3)2 = Pb2+ + NO3.

Nd(NO3)3 = Nd3+

therefore, the ions, when reduec, must be :

Nd3+ + 3 e− ⇌ Nd −2.323

Pb2+ + 2 e− ⇌ Pb(s) −0.126

Proof:

Ecell = Ecathode - Eanode = -0.126 - -2.323 = 2.197

which is the same value shown above

substitute:

Half-reaction at anode (do not multiply by factor):

The anode must contain the species tha tis being OXIDIZED, i.e. tha tloses electrons

in this case, it is Nd3+, since it has a lower reduction potential than lead

Nd3+(aq) + 3 e− ⇌ Nd(s)

Must be inverted, since this is oxidized:

Nd(s) ⇌ Nd3+(aq) + 3e-

Half-reaction at cathode (do not multiply by factor):

the cathode contains the reduction, that is, the species that is reduced, and gains electrons, it msut have the higher potential

Pb2+(aq) + 2 e− ⇌ Pb(s) −0.126

B)

calculate the Reduction potential of Nd3+ and Nd given theone of Lead is -0.126 V

As shown before

Eºcell = Ecahtode - Eanode

then,

Eºcell = 2.197V and Ecathode = -0.126

Eanode = -Eºcell + Eanode= -2.197 + (-0.126) = -2.323 V