2Br – ( aq ) + Pb 2+ ( aq ) Br 2 ( l ) + Pb( s ) 2Br – ( aq ) + Cu 2+ ( aq ) Br
ID: 926237 • Letter: 2
Question
2Br–(aq) + Pb2+(aq) Br2(l) + Pb(s)
2Br–(aq) + Cu2+(aq) Br2(l) + Cu(s)
Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq)
Cd2+(aq) + Pb(s) Cd(s) + Pb2+(aq)
Pb2+(aq) + Cu(s) Pb(s) + Cu2+(aq)
I know the answer is C, but I'm just not sure why the answer is C.
Consider the following reduction potentials:
Cd2+(aq) + 2e– Cd(s); E° = –0.40 V
Pb2+(aq) + 2e– Pb(s); E° = –0.13 V
Cu2+(aq) + 2e– Cu(s); E° = 0.34 V
Br2(l) + 2e– 2Br–(aq); E° = 1.07 V
Under standard-state conditions, which of the following reactions is spontaneous?
2Br–(aq) + Pb2+(aq) Br2(l) + Pb(s)
B.2Br–(aq) + Cu2+(aq) Br2(l) + Cu(s)
C.Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq)
D.Cd2+(aq) + Pb(s) Cd(s) + Pb2+(aq)
E.Pb2+(aq) + Cu(s) Pb(s) + Cu2+(aq)
I know the answer is C, but I'm just not sure why the answer is C.
Explanation / Answer
One whose reduction is less will as anode & the one whose reduction potential is more acts as cathode.
Always at anode oxidation & at cathode reduction takes place.
(A)2Br–(aq) + Pb2+(aq) Br2(l) + Pb(s)
standard potential of the cell , Eo = Eocathode - Eoanode
= EoPb2+/Pb - EoBr2/Br-
= -0.13 - (+1.07 ) V
= -1.20 V
(B) 2Br–(aq) + Cu2+(aq) Br2(l) + Cu(s)
standard potential of the cell , Eo = Eocathode - Eoanode
= EoBr2/Br- - EoCu2+/Cu
= +1.07 - (0.34) V
= +0.73 V
(C) Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq)
standard potential of the cell , Eo = Eocathode - Eoanode
= EoCu2+/Cu - EoCd2+/Cd
= +0.34 - (-0.40) V
= +0.74 V
(D) Cd2+(aq) + Pb(s) Cd(s) + Pb2+(aq)
standard potential of the cell , Eo = Eocathode - Eoanode
= EoPb2+/Pb - EoCd2+/Cd
= -0.13 - (-0.40) V
= +0.27 V
(E)Pb2+(aq) + Cu(s) Pb(s) + Cu2+(aq)
standard potential of the cell , Eo = Eocathode - Eoanode
= EoCu2+/Cu - EoPb2+/Pb
= +0.34 - (-0.13) V
= +0.47 V
Among these five the standard potential of (C) Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq) is +0.74V is more so it is more spontaneous.
Therefore the correct option is (C)
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