Asked by bekah
The cell shown above is a concentration cell. Both cells contain a copper solution and have copper electrodes. The only driving force for this cell is the difference in the concentration of the copper solutions. The system will react to equalize the concentration of the ions in both cells.
Oxidation: Cu (s) → Cu2+ (aq) + 2 e-
Reduction: Cu 2+ (aq) + 2 e- → Cu (s)
Determine if the following statements are True or False.
1.At the cathode, the concentration of copper (II) ions is decreasing.
2.When the cathode dissolves, copper (II) ions are formed.
3.The cell that has the lower ion concentration acts as the anode.
4.The half-cell with the higher concentration of copper (II) ions must be the cathode.
Oxidation: Cu (s) → Cu2+ (aq) + 2 e-
Reduction: Cu 2+ (aq) + 2 e- → Cu (s)
Determine if the following statements are True or False.
1.At the cathode, the concentration of copper (II) ions is decreasing.
2.When the cathode dissolves, copper (II) ions are formed.
3.The cell that has the lower ion concentration acts as the anode.
4.The half-cell with the higher concentration of copper (II) ions must be the cathode.
Answers
Answered by
bekah
the answers i have tried are F,F,T,T and F,F,T,F
Answered by
DrBob222
What do you need help with?
The lower concn will be the increasing and the higher concn will be decreasing until both have the same concn. So you know the lower concn will be oxidized and the higher concn will be reduced.
The lower concn will be the increasing and the higher concn will be decreasing until both have the same concn. So you know the lower concn will be oxidized and the higher concn will be reduced.
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