The cell shown above is a concentration cell. Both cells contain a copper soluti
ID: 879219 • Letter: T
Question
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.
True False The half-cell with the higher concentration of copper (II) ions must be the anode.
True False The cell that has the lower ion concentration acts as the cathode.
True False When the cathode dissolves, copper (II) ions are formed.
True False At the anode, the concentration of copper (II) ions is decreasing.
Explanation / Answer
All options are FALSE.
Explanation:
We can get lot of information from the given hints.
The half-cell with the higher concentration of copper (II) ions must be the anode............False.
The first hint tells us that the reduction occurs at the cathode and oxidation occurs at the anode. The second hint tells us that copper (II) ions are reactant in the half-cell where they are present at the highest concentration. This means that copper (II) ions at the highest concentration act as reactant at cathode and are reduced.
The cell that has the lower ion concentration acts as the cathode.................................False.
From the second hint we know concentration at the cathode is higher than at the anode.
When the cathode dissolves, copper (II) ions are formed.............................False.
From the first hint we know at the cathode copper (II) ions are reactants not products.
At the anode, the concentration of copper (II) ions is decreasing.............................................False.
At the anode, the concentration of copper (II) ions increases not decreases.
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