Question

In order to determine the identity of a particular transition metal (M), a voltaic cell is constructed at 25°C with the anode consisting of the transition metal as the electrode immersed in a solution of 0.018 M M(NO₃)₂, 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) <--------> M²⁺(aq) + 2e^–

Cu²⁺(aq) + 2e^– <--------> Cu(s)

The potential measured across the cell is 0.79 V. What is the identity of the metal M?

Reduction Half-Reaction	E° (V)
Cu2+(aq) + 2e– <--------> Cu(s)	+0.34
Ni2+(aq) + 2e– <--------> Ni(s)	–0.23
Cd2+(aq) + 2e– <--------> Cd(s)	–0.40
Zn2+(aq) + 2e– <--------> Zn(s)	–0.76
Mn2+(aq) + 2e– <--------> Mn(s)	–1.18

A.Ni

B.Cu

C.Mn

D.Cd

E.Zn

Answer 1

M(s) <--------> M²⁺(aq) + 2e^–                   E0   = 0.40v

Cu²⁺(aq) + 2e^– <--------> Cu(s)              E0 = 0.34v

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M(s) + Cu^2+ (aq) ----------> M^2+ (aq) + Cu(s)   E0 cell = 0.74v

E   = Ecell - 0.0592/n logQ

   = 0.74 -0.0592/2 log0.018/1

    = 0.74 -0.0296*-1.7447

    = 0.79V

M is Cd

D.Cd >>>>answer