Answer:
H2(g)+I2(s)→2HI(s)
Explanation:
Hello there!
In this case, according to the given information and unbalanced chemical reaction, we infer it must be balanced in agreement with the law of conservation of mass because the reactants side has two hydrogen and iodine atoms whereas the products side has just one. In such a way, by placing a 2 on HI, we obtain the following balanced reaction:
H2(g)+I2(s)→2HI(s)
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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 answer is decomposition, ocean release and respiration. hope this helps u :D
Answer:
0.0738 M
Explanation:
HNO3 +LiOH = LiNO3 + H2O
Number of moles HNO3 = number of moles LiOH
M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)
M(HNO3)*50.0mL = 0.100M*36.90 mL
M(HNO3) = 0.100*36.90/50.0 M = 0.0738 M
