A lone oxygen atom has 6 electrons in its outer shell which is not very stable, whereas as full octet (8 outer shell electrons) is stable. In order to achieve this two oxygen atoms will share 4 electrons, each contributing 2 electrons. Since these electrons exist within the orbitals of both atoms, to oxygen atoms essentially achieve a full octet.
Balanced Half reactions are:
At anode 2
==> Cl₂+
+ H₂O ==>
+ 2
+
At Cathode: 2
+
==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O,
,
and 
Now at Anode reaction will occur as given:
2
==> Cl₂+
+ H₂O ==>
+ 2
+
(will occur)
At Cathode:
2
+
==> H₂ (will occur)
At Cathode:
+
==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution,
will be reduced before
.
The reduction potentials for
is zero whereas reduction potential for
is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
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An atom has a nucleus and is made up of protons neutrons and electrons
Answer:
Density= 1.7g/dm3
Explanation:
Applying
P×M= D×R×T
P= 2atm, Mm= 28, D=? R= 0.082, T= 400K
2×28= D×0.082×400
D= (2×28)/(0.082×400)
D= 1.7g/dm3
Answer:
Option-4 (3:2) is the correct answer.
Explanation:
Following steps are taken to balance the given unbalanced chemical equation.
Step 1: Write the unbalanced chemical equation,
N₂ + H₂ → NH₃
Step 2: Balance Nitrogen Atoms;
There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.
N₂ + H₂ → 2 NH₃
Step 3: Balance Hydrogen Atoms;
Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.
N₂ + 3 H₂ → 2 NH₃
Now, the equation is balanced.
Step 4: Finding out mole ratios:
From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.