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.
The heat of the reaction is an extensive property: it is proportional to the quantity of the quantity that reacts.
The change in enthalpy is a measured of the heat evolved of absorbed.
When the heat is released, the change in enthalpy is negative.
The reaction of 2 moles of Na develops 368.4 kj of energy.
Calculate the number of moles of Na in 1.90 g to find the heat released when this quantity reacts.
Atomic mass of Na: 23 g/mol
#mol Na = 1.90 g / 23 g/mol = 0.0826 mol
Do the ratios: [368.4 kj/2mol ] * 0.0826 mol = 15.21 kj.
Then the answer is that 15.21 kj of heat is released (evolved)