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

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

that's what i got you can count the number of atoms

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.