Question

1. 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 MCly, 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) = M+(aq) + 3e Cu²+(aq) + 2e = Cu(3) The potential measured across the cell is 2.70 V. What is the identity of the metal? Reduction Half-Reaction Cu*(aq) + 2e Cu(s) Ce* (aq) + 3e Ce(s) Tm?"(aq) + 3e= Tm(s) Eu (aq) + 3e Eu(s) Gd"(aq) + 3e" = Gd(s) Sm(aq) + 3e Sm(s) E° (V) 0.34 -2.336 -2.319 -1.991 |--2.279 -2.304
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Answer 1

The two half reactions are Mis) = M-3 (ag) + 3e - Cu +2 (aq) + 2e - - Curso - Let Entsm and to our are the standard reduction potentials of anode and cathode respectively Multiplying equation ① by 2 and equation. @ by 3 and adding, we get a Mes) + 364+2 (aq) = RM3+ (aq) + 3&u (5) This is the overall reaction Given Futstan = 0.34 V The Standard potential of the cell is gever y po - po po celu there to and E are the reduction potentials of cathode and anode respectively

Putting the values P . 0.34 - E cert According to Nernst equation, the potential of the cell is given by t = t - 0.0591 log [m 3] n [Cutels Here n= no. of e exchanged in the - reaction which is 6 in this case. The concentrations of pure solids are taken an unity hence not included in the equation - Given Econ- 2.70V 0.04 67 M - Imita = IM forest 0.34 - Ensto Putting these values 2.10 = (0.34 - Epay -0.0591 loc (0.0467)

2.36 = - 2.36 = - E° -0.00985(-2.66) M3t/m -to to.0262 Мf3) м 2 -2.32 Eo 3t/m The to above value matches Ce 3+ + 3e Ž Ce (s) Cag) So, the metal is cel