Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
It is the boilimg point now aa
Answer: Option (3) is the correct answer.
Explanation:
Aerobic organisms are the organisms which survive and grow in the presence of oxygen.
When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.
In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.
Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.
