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.
The concentration of [H3O+] will be 6.3 x 
M
<h3>pH</h3>
Mathematically, pH = -log [H+] or -log [H3O+]
With a pH of 13.2:
-log [H3O+] = 13.2
log [H3O+] = -13.2
[H3O+] = 6.3 x M
More on pH can be found here: brainly.com/question/491373
#SPJ1
1. how does warming up help?
2. what does warming up do?
3. what would happen if you don’t warm up?
Answer:
1.
Since both components of these solutions have the same molar mass, mole fractions would be the same as mass fractions.
0.110 atm = (2/3)(Pi) + (1/3)(Pn) [1]
0.089 atm = (1/3)(Pi) + (2/3)(Pn) [2]
2*[1] - [2]:
(2)(0.110) - 0.089 atm = Pi
Pi = 0.131 atm
2*[2] - [1]:
(2)(0.089) - 0.110 atm = Pn
Pn = 0.068 atm
2.
The hydroxyl (-OH) group on the end of a longer 1-propanol molecule makes it more polar than IPA. It follows that the intermolecular forces between 1-propanol are stronger than those of IPA and thus the vapor pressure of 1-propanol should be lower than IPA.
Explanation:
