Question

In order to determine the identity of a particular lanthanide metal (M), a voltaic cell is constructed at 25°C with the anode consisting of the lanthanide metal as the electrode immersed in a solution of 0.0467 M MCl3, and the cathode consisting of a copper electrode immersed in a 1.00 M Cu(NO3)2 solution. The two half-reactions are as follows:
M(s)  M3+(aq) + 3e–


Cu2+(aq) + 2e– Cu(s)

The potential measured across the cell is 2.70 V. What is the identity of the metal?


Reduction Half-Reaction


E° (V)


Cu2+(aq) + 2e– Cu(s)


0.34


Ce3+(aq) + 3e– Ce(s)


–2.336


Tm3+(aq) + 3e– Tm(s)


–2.319


Eu3+(aq) + 3e– Eu(s)


–1.991


Gd3+(aq) + 3e– Gd(s)


–2.279


Sm3+(aq) + 3e– Sm(s)


–2.304


Tm


Gd


Ce


Eu


Sm

Answer 1

Answer:

Ce

Explanation:

From the Nernst equation:

Ecell= E° - 0.0592/n log [Red]/ [Ox]

Ecell= 2.70 V

[Red]= 1.00 M

[Ox]=0.0467 M

n= 6

2.70= E° - 0.0592/6 log 1.00/0.0467

2.70= E° - 0.013

E°= 2.70 + 0.013

E°= 2.713 V

But E°= E° cathode - E°anode

E° cathode= +0.34 V

E°anode= 0.34-2.713

E°anode= -2.373 V