E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
Answer:
The Barium flame is green because it is a difficult flame to excite, therefore for it to trigger a flame it is necessary that it be too excited for it to occur.
The reddish color of calcium is due to its high volatility and it is sometimes very difficult to differentiate it from strontium.the compression of these elements is due to being able to make them work during combustion
Explanation:
The flame test is a widely used qualitative analysis method to identify the presence of a certain chemical element in a sample. To carry it out you must have a gas burner. Usually a Bunsen burner, since the temperature of the flame is high enough to carry out the experience (a wick burner with an alcohol tank is not useful). The flame temperature of the Bunsen burner must first be adjusted until it is no longer yellowish and has a bluish hue to the body of the flame and a colorless envelope. Then the tip of a clean platinum or nichrome rod (an alloy of nickel and chromium), or failing that of glass, is impregnated with a small amount of the substance to be analyzed and, subsequently, the rod is introduced into the flame, trying to locate the tip in the least colored part of the flame.
The electrons in these will jump to higher levels from the lower levels and immediately (the time that an electron can be in higher levels is of the order of nanoseconds), they will emit energy in all directions in the form of electromagnetic radiation (light) of frequencies characteristics. This is what is called an atomic emission spectrum.
At a macroscopic level, it is observed that the sample, when heated in the flame, will provide a characteristic color to it. For example, if the tip of a rod is impregnated with a drop of Ca2 + solution (the previous notation indicates that it is the calcium ion, that is, the calcium atom that has lost two electrons), the color observed is brick red .
Answer:
Oxygen Gas
Explanation:
The balanced equation shows us the reactant ratio of the reaction.
This means that for every one mole of CH3CH2OH, we need 3 moles of O2 to react with it. Because we need more O2, (3x as much) than ethanol and we have the same given amount (1 mole of each), the oxygen will be the limiting reagent. (1 mole of oxygen would only require 1/3 moles of ethanol to react).
Hope this helped!
Look at the liter man it’s a great way to learn how much every liquid measurement is
