Answer:a
Explanation:
i think so cs i read the ariticle
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:
Explanation:
Ketcher 01232019462D 1 1.00000 0.00000 0 5 4 0 0 0 999 V2000 -0.0330 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 2.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 3.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 1.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 3 1 0 0 0 2 4 1 0 0 0 3 5 1 0 0 0 M END
Na + NaNO3 = Na2O + N2
4 Na + 2 NaNO3 = 6 Na2O + N2
6 Na on each side
2 N on each side
6 O on each side
