Homework SoursePlease help me to solove this question step by step I need t
Please help me to solove this question step by step, I need to see all the details. Credits will not be given if there is only
answer provided (The correct answer is Eu, but I don\'t know how to get it)
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 0.0136 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. The
potential measured across the cell is 2.37 V. What is the identity of the metal?
M(s) à M3+(aq) + 3e–
Cu2+(aq) + 2e–à Cu(s)
Reduction Half-Reaction
â„° ° (V)
Reduction Half-Reaction
â„° ° (V)
Cu2+(aq) + 2e–à Cu(s)
0.34
Eu3+(aq) + 3e–à Eu(s)
–1.991
Ce3+(aq) + 3e–à Ce(s)
–2.336
Gd3+(aq) + 3e–à Gd(s)
–2.279
Dy3+(aq) + 3e–à Dy(s)
–2.295
Sm3+(aq) + 3e–à Sm(s)
–2.304
a) Ce
b) none of these
c) Dy
d) Gd
e) Eu
| Reduction Half-Reaction | â„° ° (V) | Reduction Half-Reaction | â„° ° (V) | |
| Cu2+(aq) + 2e–à Cu(s) | 0.34 | Eu3+(aq) + 3e–à Eu(s) | –1.991 | |
| Ce3+(aq) + 3e–à Ce(s) | –2.336 | Gd3+(aq) + 3e–à Gd(s) | –2.279 | |
| Dy3+(aq) + 3e–à Dy(s) | –2.295 | Sm3+(aq) + 3e–à Sm(s) | –2.304 |
Solution
emf = emf0 + (.059 log[M3+]^2/[cu+2]^3)/n
n = 6
2.64 = emf0 + .33
emfo = 2.31
here = emfo = 1.991 +.34 = 2.331
