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
M(Ag)=12.5 g
Nₐ=6.022*10²³ mol⁻¹
n(Ag)=m(Ag)/M(Ag)
N=n(Ag)*Nₐ
N=Nₐm(Ag)/M(Ag)
N=6.022*10²³mol⁻¹*12.5g/107.868g/mol=6.97*10²²
A. 6.97 × 10²² atoms
<h3><u>Answer;</u></h3>
The value of Kc does not depend on starting concentrations.
<h3><u>Explanation;</u></h3>
- <em><u>At constant temperature, changing the equilibrium concentration does not affect the equilibrium constant, because the rate constants are not affected by the concentration changes. </u></em>
- When the concentration of one of the participants is changed, the concentration of the others vary in such a way as to maintain a constant value for the equilibrium constant.
I believe the answer is C because the first two are mechanical hazards and C mentions plug which is electrical.
