32. Consider the following two standard reduction potentials: Pt2"(aq) + 2 e → Pt (s); E° = 1.180 V Pb"(aq) + 2 e → Pb(s); E° = –0.130 V Part 1. Which of the following is the balanced overall reaction and standard cell potential of a working electrochemical (or voltaic) cell constructed from half-cells with the given half reactions? Note that "working" means that the electrochemical reaction is spontaneous. a) Pt2"(aq) + Pb(s) → Pt(s) + Pb²"(aq); = 1.310 V cell Pt(s) + Pb* (aq) – Pt"(aq) + Pb(s); cell - 1.310 V Pt* (aq) + Pb3 (aq) → Pt(s) + Pb(s); cell = 1.050 V d) p Pr*"(aq) + Pb(s) → Pt(s) + Pb²"(aq); cell 0.655 V e) Pt(s) + Pb²"(aq) → Pt2"(aq) + Pb(s); £cel 0.655 V f) none of the above

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32. Consider the following two standard reduction potentials:
Pt2"(aq) + 2 e → Pt (s); E° = 1.180 V
Pb2"(aq) + 2 e
→ Pb(s); E° = -0.130 V
%3D
Part 1. Which of the following is the balanced overall reaction and standard cell potential of a working electrochemical (or
voltaic) cell constructed from half-cells with the given half reactions? Note that "working" means that the
electrochemical reaction is spontaneous.
a)
" Pt2 (aq) + Pb(s) → Pt(s) + Pb²*(aq); Ecelnl= 1.310 V
+ Pb(s) → Pt(s) + Pb²*(aq); £
%3D
b)
Pt(s) + Pb²*(aq) → Pt2*(aq) + Pb(s);
E
cell -1.310 V
c)
Pt2"(aq) + Pb²*(aq)→ Pt(s) + Pb(s);
= 1.050 V
%3D
->
cell
d)
" Pr2"(aq) Een=0.655 V
+ Pb(s) –
Pt(s) + Pb²*(aq);
%3D
cell
e)
Pt(s) + Pb²"(aq) → Pt2*(aq) + Pb(s); £
cell=-0.655 V
f) none of the above
Transcribed Image Text:32. Consider the following two standard reduction potentials: Pt2"(aq) + 2 e → Pt (s); E° = 1.180 V Pb2"(aq) + 2 e → Pb(s); E° = -0.130 V %3D Part 1. Which of the following is the balanced overall reaction and standard cell potential of a working electrochemical (or voltaic) cell constructed from half-cells with the given half reactions? Note that "working" means that the electrochemical reaction is spontaneous. a) " Pt2 (aq) + Pb(s) → Pt(s) + Pb²*(aq); Ecelnl= 1.310 V + Pb(s) → Pt(s) + Pb²*(aq); £ %3D b) Pt(s) + Pb²*(aq) → Pt2*(aq) + Pb(s); E cell -1.310 V c) Pt2"(aq) + Pb²*(aq)→ Pt(s) + Pb(s); = 1.050 V %3D -> cell d) " Pr2"(aq) Een=0.655 V + Pb(s) – Pt(s) + Pb²*(aq); %3D cell e) Pt(s) + Pb²"(aq) → Pt2*(aq) + Pb(s); £ cell=-0.655 V f) none of the above
Part 2. In the above working electrochemical (or voltaic) cell in Part 1, what species is reduced in cathode?
a) Pt (s)
b) Pt2* (aq)
c) Pb (s)
d) Pb²* (aq)
e) None of the above
Part 3. From Part 1, one can conclude that
a) Pt (s) is a stronger reducing agent than Pb (s).
b) Pb (s) is a stronger reducing agent than Pt (s).
c) Pt (s) is a stronger oxidizing agent than Pb (s).
d) Pb (s) is a stronger oxidizing agent than Pt (s).
e) none of the above
Transcribed Image Text:Part 2. In the above working electrochemical (or voltaic) cell in Part 1, what species is reduced in cathode? a) Pt (s) b) Pt2* (aq) c) Pb (s) d) Pb²* (aq) e) None of the above Part 3. From Part 1, one can conclude that a) Pt (s) is a stronger reducing agent than Pb (s). b) Pb (s) is a stronger reducing agent than Pt (s). c) Pt (s) is a stronger oxidizing agent than Pb (s). d) Pb (s) is a stronger oxidizing agent than Pt (s). e) none of the above
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