1. · Involved reactions:
2.     Cu²⁺ + 2·e^– → Cu, z = 2
3.     6·H₂O → 4·H₃O⁺ + O₂ + 4·e^– ; z = 4
4. · m/M = I·t/(z·F)
5.     t = F·m·z/(I·M)
6.     t = 96485·0.8·2/(0.18·63.5)
7.     t = 13506 s = about 4 h
8. · m/M = n(O₂) = I·t/(z·F)
9.     n(O₂) = 0.18·13506/(4·96485) = 0.00629 mol
10. · p·V = n·R·T
11.     V(O₂)= n·R·T/p
12.     V(O₂) = 0.00629·8.314·298/101300
13.     V(O₂) = 0.000154 m³ = approx. 154 ml
14. · Units: [mol·J/(mol·K) ·K / N/m²]

800 mg of copper at I=180mA should be removed from a copper sulphate solution at the cathode. Calculate (a) the time duration of electrolysis and (b) determines the volume of oxygen that is created simultaneously under normal conditions (298 K, 101300 N/m²).
The general gas equation is p·V=n·R·T, whereas R = 8.314 J/molK.
Note : identical exercise also at the same time, subjects salts.

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1. · Copper sulphate solution: CuSO₄, or Cu²⁺/SO2-4
2.     Cu²⁺ + 2e^– → Cu, z = 2
3. · m/M = I·t/(z·F)
4.     m = I·t·M/(z·F) = 5.8·3600·3·63.5/(2·96485) = 20.6 g
5.     Units: [A·s·g/mol/(A·s/mol)] = [g]
6. · Es of which: approximate: the volumes (!) the bullet layer is equal the surface of the sphere multiplied by with the height, or Thick (:d) the layers V = 4·π·r²·d
7. · Density = m/V
8.     m = density · V = Dichte·4·π·r²·d
9.    
10.     → d = m/(Dichte·4·π·r²)
11.     d = 20.63/(8.9·4·π·8.3²) = 2.67·10^-3cm = 0.026 mm

Calculate the thickness of the copper layer that a current of 5.8 A dissolves evenly in 3 hours on a sphere with a radius of 8.3 cm.
Note 1: The surface of a sphere is 4·pi·r²
Note 2: Coppering was carried out with a copper sulphate solution
Note 3: Dichte(Cu) = 8.9 g/cm³
Note : identical exercise also at the same time, subjects salts.

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1. · The volume of the surface to be silted shall be:
2.     V= O·h = 120cm²·0.02cm = 2.4cm³
3. · With density = m/V:
4.     m=thickness·V= 10.5/cm³ · 2.4 cm³ = 25.2 grams
5. · Silver nitrate: AgNO₃, i.e. Ag⁺/NO–3
6.     Ag⁺ + 1e^– → Ag, z= 1
7. · m/M = I·t/(z·F)
8.     t = m·z·F/(M·I) = 25.2·1·96485/(107.9·10)
9.     t = 2253 s , about 40 minutes

A shell with a surface of 120 cm² should through electrolysis a silver nitrate solution a 0.2 mm thick silver transhipment received. Calculate the time required if the current strength is 10 A. The density of the silver is 10.5 g/cm³.
Note : identical exercise also at the same time, subjects salts.

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1. · 10 A = 10 C/s
2. · Number of electrons per second:
3.     10/1.602·10^-19 = 6.24·10¹⁹
4. · Total Masse(e^–) per second:
5.     6.24·10¹⁹·9.1E.31 = 5.68·10^-11 kg
6. · How long to a total of 1 kg:
7.     1/5.68·10^-11 = 1.76·10¹⁰ s = about 558 years

How long does a current of 10 A have to flow (about the usual 'household current ') before electrons of a total mass of 1 kg are passed through the conductor cross-section?
Notes m(e^–) = 9.1·10^-31 kg, load 1e^– = 1.602·10^-19 C
Note : identical exercise also at the same time, subjects salts.

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1. CuCl₂: Ions Cu²⁺ or Cl^–
       Cu²⁺ + 2e^– → Cu, z = 2
2. m/M = I·t/(z·F)
       t = m·z·F/(M·I) = 3.5·2·96485/(63.546·0.45)
       t = 23618 s = about 6.6 h

Note: 35 µm correspond to 35·10^-6m

1. The volume of the chromium layer to be added shall be:
       V= O·h = 15m^2·35·10-6m = 0.000525 m³ = 525 cm³
2. With density = m/V:
       m=density·V= 7.15g/cm³ · 525 cm³ = 3753.75 grams
3. CrO₃, participants ions: Cr⁶⁺/O^2-
       Cr⁶⁺ + 6e^– → Cr, z = 6
4. m/M = I·t/(z·F)
       t = m·z·F/(M·I) = 96485·6·3753/(52·450)
       t = 92848 s (at 100% efficiency)
5. Efficiency of 90%: 103164 seconds