ManganeseMnO, + 4H* + 3e¯ =MnO2(s) + 2H2OMn+ + e= Mn²+Mno, + 8H* + 5e¯ = Mn²+ + 4H2OMn2O3(s) + 6H* + 2e¯= 2MN²* + 3H,OMnO,(s) + 4H* + 2e¯=Mn²+ + 2H,OMn(EDTA)¯ + e¯=Mn(EDTA)²-MnO, + e¯ = MnO3Mn,O3(s) + H,0 + 2e¯=2Mn;O,(8) + 20H¯Mn;04(s) + 4H;0 + 2e¯ =3Mn(OH)2(s) + 20H"1.692-0.6711.561.81.507-0.646-0.9261.4851.230-0.6090.825-1.100.560.002-0.352-2.05-1.256-1.61Mn2+Mn(OH)2(8) + 2e¯=Mn(s) + 20H¯- 2e¯ =Mn(s)-1.182-1.129-1.565-1.10CopperCu* + e= Cu(s)Cu²+ + 2e¯ = Cu(s)Cu²+ + e Cu"CuCl(s) + e¯= Cu(s) + Cl¯Cu(IO3)2(s) + 2e¯=Cu(s) + 2IO,Cu(ethylenediamine), + e¯=Cu(s) + 2 ethylenediamineCuI(s) + e¯= Cu(s) + I¯Cu(EDTA)²- + 2e¯=Cu(s) + EDTA“-Cu(OH)2(s) + 2e¯=Cu(s) + 20H'Cu(CN), + e¯= Cu(s) + 2CNCuCN(s) + e¯=Cu(s) + CN0.5180.3390.1610.137-0.079-0.119-0.185-0.216-0.222-0.429-0.639BromineBro, + 2H* + 2e¯=BrO, + H2OHOBR + H* + e¯=}Br2(1) + H,0Bro, + 6H* + 5e¯ =Br>(1) + 3H2OBr2(aq) + 2e¯ = 2BrBr,(1) + 2e¯= 2BrBr, + 2e= 3BrBro + H2O + 2e¯=Br + 20H™Bro, + 3H,0 + 6e¯=Br + 60H1.745-0.5111.584-0.751.5131.0981.0781.0620.766-0.419-0.499-0.611-0.512-0.940.613-1.287 Consider the following cell:Mn(s) | MnBr2(0.046 M) || CuBr(0.024 M) | Cu(s)a) Using the standard reduction potentials table, search for the appropriate half reactions that add to the reaction of this full cell. What are the values of the standard potentials of the half cells?E°, =VE°- =b) Calculate the standard potential of the full cell.E° =Vc) Calculate the equilibrium constant of the full cell.K =x 10^

In this question, we have to identify the cell 1-oxidation half cell 2-reduction half cell…

Consider the following cell: Mn(s) | MnBr2(0.046 M) || CuBr(0.024 M) | Cu(s) a) Using the standard reduction potentials table, search for the appropriate half reactions that add to the reaction of this full cell. What are the values of the standard potentials of the half cells? E°, = E°. = V b) Calculate the standard potential of the full cell. E° = V c) Calculate the equilibrium constant of the full cellI. K= x 10A

Consider the following cell: Mn(s) | MnBr2(0.046 M) || CuBr(0.024 M) | Cu(s) a) Using the standard reduction potentials table, search for the appropriate half reactions that add to the reaction of this full cell. What are the values of the standard potentials of the half cells? E°, = E°. = V b) Calculate the standard potential of the full cell. E° = V c) Calculate the equilibrium constant of the full cellI. K= x 10A

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

Manganese
MnO, + 4H* + 3e¯ =MnO2(s) + 2H2O
Mn+ + e= Mn²+
Mno, + 8H* + 5e¯ = Mn²+ + 4H2O
Mn2O3(s) + 6H* + 2e¯= 2MN²* + 3H,O
MnO,(s) + 4H* + 2e¯=Mn²+ + 2H,O
Mn(EDTA)¯ + e¯=Mn(EDTA)²-
MnO, + e¯ = MnO
3Mn,O3(s) + H,0 + 2e¯=2Mn;O,(8) + 20H¯
Mn;04(s) + 4H;0 + 2e¯ =3Mn(OH)2(s) + 20H"
1.692
-0.671
1.56
1.8
1.507
-0.646
-0.926
1.485
1.230
-0.609
0.825
-1.10
0.56
0.002
-0.352
-2.05
-1.256
-1.61
Mn2+
Mn(OH)2(8) + 2e¯=Mn(s) + 20H¯
- 2e¯ =Mn(s)
-1.182
-1.129
-1.565
-1.10
Copper
Cu* + e= Cu(s)
Cu²+ + 2e¯ = Cu(s)
Cu²+ + e Cu"
CuCl(s) + e¯= Cu(s) + Cl¯
Cu(IO3)2(s) + 2e¯=Cu(s) + 2IO,
Cu(ethylenediamine), + e¯=Cu(s) + 2 ethylenediamine
CuI(s) + e¯= Cu(s) + I¯
Cu(EDTA)²- + 2e¯=Cu(s) + EDTA“-
Cu(OH)2(s) + 2e¯=Cu(s) + 20H'
Cu(CN), + e¯= Cu(s) + 2CN
CuCN(s) + e¯=Cu(s) + CN
0.518
0.339
0.161
0.137
-0.079
-0.119
-0.185
-0.216
-0.222
-0.429
-0.639
Bromine
Bro, + 2H* + 2e¯=BrO, + H2O
HOBR + H* + e¯=}Br2(1) + H,0
Bro, + 6H* + 5e¯ =Br>(1) + 3H2O
Br2(aq) + 2e¯ = 2Br
Br,(1) + 2e¯= 2Br
Br, + 2e= 3Br
Bro + H2O + 2e¯=Br + 20H™
Bro, + 3H,0 + 6e¯=Br + 60H
1.745
-0.511
1.584
-0.75
1.513
1.098
1.078
1.062
0.766
-0.419
-0.499
-0.611
-0.512
-0.94
0.613
-1.287

Consider the following cell:
Mn(s) | MnBr2(0.046 M) || CuBr(0.024 M) | Cu(s)
a) Using the standard reduction potentials table, search for the appropriate half reactions that add to the reaction of this full cell. What are the values of the standard potentials of the half cells?
E°, =
V
E°- =
b) Calculate the standard potential of the full cell.
E° =
V
c) Calculate the equilibrium constant of the full cell.
K =
x 10^

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