Answer:
Amount of salt needed is around 2.3*10³ g
Explanation:
The salt content in sea water = 3.5 %
This implies that there is 3.5 g salt in 100 g sea water
Density of seawater = 1.03 g/ml
Volume of seawater = volume of tank = 62.5 L = 62500 ml
Therefore, the amount of seawater required is:
The amount of salt needed for the calculated amount of seawater is:
Zinc would be considered the strongest reducing agent.
<h3>Reducing agent</h3>
A reducing agent is a chemical species that "donates" one electron to another chemical species in chemistry (called the oxidizing agent, oxidant, oxidizer, or electron acceptor). Earth metals, formic acid, oxalic acid, and sulfite compounds are a few examples of common reducing agents.
Reducers have excess electrons (i.e., they are already reduced) in their pre-reaction states, whereas oxidizers do not. Usually, a reducing agent is in one of the lowest oxidation states it can be in. The oxidation state of the oxidizer drops while the oxidizer's oxidation state, which measures the amount of electron loss, increases. The agent in a redox process whose oxidation state rises, which "loses/donates electrons," which "oxidizes," and which "reduces" is known as the reducer or reducing agent.
Learn more about reducing agent here:
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<h3 />
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.
