The cathode of an electrolysis cell contains (ml75) an internal solution of (0.1 M) of copper ions (2 + Cu) in addition to the copper electrode. The zinc electrode was immersed in a solution of (30) ml of (0.03M) of zinc, which acts as an anode. Current (0.02 ampere cm2). The cell impedance was (5 ohm). (0.15.V Cu → Cu 0.34 + = EO Zn + 2 + 2e → Zn V0.76 + = EO Calculate 1- Anode Voltage 2 - General Cell Voltage Without Influence Current 3- Primary Current Jo 9:05

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Chapter1: Units, Trigonometry. And Vectors
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The cathode of an electrolysis cell contains
(ml75) an internal solution of (0.1 M) of
copper ions (2 + Cu) in addition to the copper
electrode. The zinc electrode was immersed in
a solution of (30) ml of (0.03M) of zinc, which
acts as an anode. Current (0.02 ampere cm2).
The cell impedance was (5 ohm). (0.15.V Cu
+ 2 + 2e → Cu 0.34 + = EO Zn + 2 + 2e → Zn
V0.76 + = EO Calculate 1- Anode Voltage 2 -
General Cell Voltage Without Influence Current
3- Primary Current
->
Jo 9:05 /
Transcribed Image Text:The cathode of an electrolysis cell contains (ml75) an internal solution of (0.1 M) of copper ions (2 + Cu) in addition to the copper electrode. The zinc electrode was immersed in a solution of (30) ml of (0.03M) of zinc, which acts as an anode. Current (0.02 ampere cm2). The cell impedance was (5 ohm). (0.15.V Cu + 2 + 2e → Cu 0.34 + = EO Zn + 2 + 2e → Zn V0.76 + = EO Calculate 1- Anode Voltage 2 - General Cell Voltage Without Influence Current 3- Primary Current -> Jo 9:05 /
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