redox, Redux table, It’s not funny 1 (Redox-30) How is The Other goods of heading 8401 Equation ? How is The Other goods of heading 8401 Equation for diluted LManufacture in which all the materials used are: of the equal to Flat-rolled products of iron or steel Given be One concentrated Cu2+-Lösung and One diluted Cu2+-Lösung, in which in each case One Cu-Electrodes Rags. LManufacture in which all the materials used are:, It 's just … Separated by: a For the purposes of this Regulation:, with each other Connected are? Use of which: other The concept Reduction, Oxidation, Court of Justice to The Catheter Responsibility. Anode. How Calculated The Commission The Tensions between two Different Half-cells ? Calculate This Potential One of them Silver cell with The Concentration of 0.1 mol/l Calculate This Potential One of them Silver cell with The Concentration of 0.01 mol/l How Large is The Tensions between two Semicells with The Tensions 0.1 Responsibility. 0.01 mol/L One of them. Zinc half cell will be to The Factor 1000 diluted. Calculate The Voltage decrease Responsibility. This new Potential.

  Aufgabe: Redox-30    

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1. How is The Other goods of heading 8401 Equation ?
2. How is The Other goods of heading 8401 Equation for diluted LManufacture in which all the materials used are: of the equal to Flat-rolled products of iron or steel
3. Given be One concentrated Cu²⁺-Lösung and One diluted Cu²⁺-Lösung, in which in each case One Cu-Electrodes Rags. LManufacture in which all the materials used are:, It 's just … Separated by: a For the purposes of this Regulation:, with each other Connected are? Use of which: other The concept Reduction, Oxidation, Court of Justice to The Catheter Responsibility. Anode.
4. How Calculated The Commission The Tensions between two Different Half-cells ?
5. Calculate This Potential One of them Silver cell with The Concentration of 0.1 mol/l
6. Calculate This Potential One of them Silver cell with The Concentration of 0.01 mol/l
7. How Large is The Tensions between two Semicells with The Tensions 0.1 Responsibility. 0.01 mol/L
8. One of them. Zinc half cell will be to The Factor 1000 diluted. Calculate The Voltage decrease Responsibility. This new Potential.

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Lösung:

What ? This is happening, where: The

1. E = E^o + 0.059/n · log(c^a(Oxidierte Form)/ c^b(Reduzierte Form)
2. E = E^o + 0.059/n · log(c(verdünnte Lösung)/c(konzentrierte Lösung))
3. Finally must be The Both LManufacture in which all the materials used are: The equal to Concentration have. Therefore,:
   · The Concentration must In the case of The higher Concentration Lthe island decrease. The only Get out of here . would be, that The Copper ions to elementary Copper will be (Cu²⁺ → Cu), It 's must So what do you mean? One Reduction to take place, According to the definition to The Catheter.
   · Conversely Applicable, that The deeper Concentration The Commission increase must. This Goes It 's just … This is it., that fixed Copper (from The Electrodes) gained will be, Cu → Cu²⁺, (Oxidation), which in The Lthe island Going. The Oxidation process finds to The Anode Instead of.
4. ΔE = E_{Kathodenreaktion} – E_{Anodenreaktion}
5. Ag⁺ + e^– → Ag, E^o=0.8 V, z=1
       · Reduced Form Ag, c(Ag) = 1
       · Oxidated Form Ag⁺ , c(Ag⁺)= 0.1 mol/L
       · E₁ = 0.8 + 0.059/1 · log(0.1/1)= 0.741 V
       · or .. E₁ = 0.8 – 0.059/1 · log(1/0.1)= 0.741 V
6. Ag⁺ + e^– → Ag, E^o=0.8 V, z=1
       · Reduced Form Ag, c(Ag) = 1 mol/L
       · Oxidated Form Ag⁺ , c(Ag⁺)= 0.01 mol/L
       · E₂ = 0.8 + 0.059/1 · log(0.01/1)= 0.682 V
       · or .. E₂ = 0.8 – 0.059/1 · log(1/0.01)= 0.682 V
7. From The previously calculations is given The Commission:
       · [Ag⁺]= 0.1 mol/L → E₁=0.741 V
       · [Ag⁺]= 0.01 mol/L → E₂=0.682 V
       · ΔE = E₁ – E₂ = 0.06 V
8. Zn → Zn²⁺ + 2e^–, E^o=-0.76 V, z=2
       · Reduced Form Zn, c(Zn) = 1 mol/L
       · Oxidated Form Zn²⁺ , c(Zn²⁺)= 0.001 mol/L
       · E = E^o + 0.059/2 · log(0.001/1) = -0.849 V

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