Indicate what happens to the concentration of Pb2+ in each half cell
1 answer:
Answer:
Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below
The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.
Explanation:
The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.
In the Left Beaker (Left half cell), their is less concentration
Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side
In the Right Beaker (right half cell), their is more concentration
Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side
Electrons will flow from Left to Right direction.
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Answer: 1. moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove
moles of
2.
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by = moles of sodium iodide.
Mass of sodium iodide= (1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .