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
Answer:
(A) The shorter the wavelength, the more total energy the wave contains.
(B) The longer the wavelength, the less total energy the wave contains.
Explanation:
The wavelength (λ), frequency (f) and energy (E) are interrelated. This relationship between them is represented in the following equations:
λ = v/f and E = hf
Where;
λ = wavelength (m)
f = frequency (Hz)
E = energy (Joules)
v and h represents speed of light and Planck's constants respectively.
Combining both equations, E = hc/λ
This equation shows that ENERGY (E) is directly proportional to the frequency (f) but inversely proportional to the wavelength (λ). This means that "the shorter the wavelength, the more total energy a wave contains" and vice versa.
However, the higher the frequency, the more the total energy the wave contains and vice versa.
