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:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
Answer:
94.61 %
Explanation:
percent yield = (actual yield / theoretical yield) X 100%
The balanced equation for the reaction is:
H2 (g ) + Cl2(g) => 2 HCl (l)
So, the theoretical yield =
7.25g of chlorine X (2mol of Cl / 35.453 g of Cl) X (2mol Cl / 2mol of HCl) X 37.469g of Hcl / 2mol of Hcl = 0.409 x 18.735 = 7.663g of Hcl
Using this theoretical yield and the provided value for actual yield, the percent yield = (actual yield / theoretical yield) X 100%
= (7.25 g / 7.663g) X 100
= 94.61 %
Answer:
2l- ---> l2 + 2e- is the anode
2H+ + 2e- ---> H2(g) is the cathode
Explanation:
Oxidation occurs when a metal loses two or more electrons in a redox chemical reaction and reduction is when it gains. Thus, oxidation is the anode and reduction is the cathode.
