63. Which of the following metal cations is the best oxidizing agent? 124. A 2.4

Question

63. Which of the following metal cations is the best oxidizing agent?

124. A 2.4610?2L sample of a solution of Cu + requires 3.2310?2L of 0.132M KMnO4 solution to reach the equivalence point. The products of the reaction are Cu2+ and Mn2+

What is the concentration of the Cu2+solution?

[Cu2+] =???????????? M

82. A Cu/Cu2+ concentration cell has a voltage of 0.24V at 25 ?C. The concentration of Cu2+ in one of the half-cells is 1.410?3M .

What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell to be the lower of the two concentrations.)

[Cu2+] =???????????? M

75. An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)?Pb2+(aq, 0.26M )+2e?
Red: MnO?4(aq, 1.25M )+4H+(aq, 1.8M )+3e??
MnO2(s)+2H2O(l)

Compute the cell potential at 25 ?C.

Please answer all 4 quesitions (and show how) in order to get the 750 points...thank you :)

Explanation / Answer

1) the best oxidizing agent should have more positive reduction potential .

So Cu+2 is the best oxidizig agent .

2) the reaction is given by

Mn+7 + 5Cu+ -------> Mn+2 + 5Cu+2

at equivalnce point

molarity of Cu+ x volume of Cu+ = 5 x molarity of KMn04 x volume of KMn04

molarity of Cu+ x 2,46 x 10-2 = 5 x 0.132 x 3.23 x 10-2

molarity of Cu+ =0.866 M

moles of Cu + 0.866 x 2.46 x 10-2

moles of Cu+ = 0.021318

moles of Cu+2 = moles of Cu+

moles of Cu+2 = 0.021318

total volume = 5.69 x 10-2

conc of Cu+2 = 0.021318 / 5.69 x 10-2

conc of Cu+2 = 0.3746

so the conc of Cu+2 is 0.3746 M

3) for a concnetration cell

Eo= 0

so E = Eo - ( 0.059/n ) log [ anode / cathode ]

the reaction is

Cu + Cu+2 -----> Cu+2 + Cu

so 0.24 = 0 - ( 0.0591 /2 ) log [ anode / cathode ]

log [ anode / cathode ] = -8.12

log [ anode / 1.4 x 10-3 ] = -8.12

[ anode ] = 1.05 x 10-11

the concentration in other half cell is 1.05 x 10-11 M

4)

the final reaction is given by

3 Pb + 2 Mn04- + 8H+ ----> 3 Pb+2 + 2Mn02 + 4H20

Eo = Eo cathode - Eo anode

Eo = 1.70 - ( -0.13 )

Eo = 1.83 V

E = Eo - ( 0.0591 / n ) log [ anode / cathode ]

E = 1.83 - ( 0.0591 / 6 ) log [ ( Pb+2 ) ^3 / ( Mn04-) ^2 ( H+) ^8 )

E= 1.83 - ( 0.0591 /6 ) log [ ( 0.26 ) ^3 / ( 1.25 ) ^2 ( 1.8 ) ^8 )

E =1.87 V

so the cell potential is 1.87 V

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