Redox, Redox table, Nernst 1 (Redox-30)

1. What 's the Nernst equation ?
2. What is the Nernst equation for diluted solutions of the same metal
3. A concentrated Cu²⁺ – solution and a diluted Cu²⁺ – solution are given, each containing a Cu electrode. Solutions, only separated by a diaphragm, connected to each other? Use of which: other the concept reduction, oxidation, court of Justice at the catheter or [translation unavailable].
4. How do you calculate the voltage between two different half-cells ?
5. Calculate the potential of a silver cell with a concentration of 0.1 mol/l
6. Calculate the potential of a silver cell with a concentration of 0.01 mol/l
7. What is the voltage between two half-cells with a voltage of 0.1 and 0.01 mol/L?
8. A zinc half cell is diluted by a factor of 1000. Calculate the voltage drop and the new potential.

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• October 16, 2026 at 4:42 am #9508
    Exercise: Redox-30    

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  1. What 's the Nernst equation ?
  2. What is the Nernst equation for diluted solutions of the same metal
  3. A concentrated Cu²⁺ – solution and a diluted Cu²⁺ – solution are given, each containing a Cu electrode. Solutions, only separated by a diaphragm, connected to each other? Use of which: other the concept reduction, oxidation, court of Justice at the catheter or [translation unavailable].
  4. How do you calculate the voltage between two different half-cells ?
  5. Calculate the potential of a silver cell with a concentration of 0.1 mol/l
  6. Calculate the potential of a silver cell with a concentration of 0.01 mol/l
  7. What is the voltage between two half-cells with a voltage of 0.1 and 0.01 mol/L?
  8. A zinc half cell is diluted by a factor of 1000. Calculate the voltage drop and the new potential.

  ---

  
  

    Exercise: Redox-30    

  ⓘ    🔄 Reload     🖨️ print           ◀  –   ▶     ?! Question/Answer ?!

  1. What 's the Nernst equation ?
  2. What is the Nernst equation for diluted solutions of the same metal
  3. A concentrated Cu²⁺ – solution and a diluted Cu²⁺ – solution are given, each containing a Cu electrode. Solutions, only separated by a diaphragm, connected to each other? Use of which: other the concept reduction, oxidation, court of Justice at the catheter or [translation unavailable].
  4. How do you calculate the voltage between two different half-cells ?
  5. Calculate the potential of a silver cell with a concentration of 0.1 mol/l
  6. Calculate the potential of a silver cell with a concentration of 0.01 mol/l
  7. What is the voltage between two half-cells with a voltage of 0.1 and 0.01 mol/L?
  8. A zinc half cell is diluted by a factor of 1000. Calculate the voltage drop and the new potential.

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  October 16, 2026 at 4:42 am #9509
  Solution:

  What happens if the

  1. E = E^o + 0.059/n · log(c^a(Oxidierte [translation unavailable])/ c^b(Reduzierte [translation unavailable])
  2. E = E^o + 0.059/n · log(c(verdünnte Lösung)/c(konzentrierte Lösung))
  3. Finally, the two solutions must have the same concentration.
     · The concentration must decrease at the higher concentration solution. Would be, that the copper ions to elementary copper are (Cu²⁺ → Cu), it must thus a reduction to take place, according to the definition at the catheter.
     · Conversely, the deeper concentration must be increased. This goes only so, that fixed copper (from the electrodes) gained is, Cu → Cu²⁺, (oxidation), which in the solution going. The oxidation process takes place at the anode.
  4. ΔE = E_{Kathodenreaktion} – E_{Anodenreaktion}
  5. Ag⁺ + e^– → Ag, E^o=0.8 V, z=1
         · Reduced form Ag, c(Ag) = 1
         · Oxidated [translation unavailable] 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 [translation unavailable] 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. It follows from the previous calculations:
         · [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 [translation unavailable] Zn²⁺ , c(Zn²⁺)= 0.001 mol/L
         · E = E^o + 0.059/2 · log(0.001/1) = -0.849 V

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