The cell notation for the voltaic cell that incorporates the redox reaction Mg(s) + Sn²⁺(aq) → Mg²⁺(aq) + Sn(s) is Mg(s)|Mg²⁺(aq)║Sn²⁺(aq)|Sn(s).
The cell notation for a voltaic cell is the following:
anode ║ cathode
The anode is where the oxidation reaction takes place and the cathode is where the reduction happens.
The given reaction is:
Mg(s) + Sn²⁺(aq) → Mg²⁺(aq) + Sn(s)
We can see that <u>magnesium </u>is <u>oxidizing</u> (it is losing electrons) and that <u>tin </u>is <u>reducing</u> (it is gaining electrons).
These two processes can be represented in the following half-reactions:
Oxidation: Mg(s) → Mg²⁺(aq)
Reduction: Sn²⁺(aq) → Sn(s)
Which in <u>cell notation</u> is:
Anode: Mg(s)|Mg²⁺(aq)
Cathode: Sn²⁺(aq)|Sn(s)
Hence, the <u>notation</u> for the <u>voltaic cell</u> is:
Mg(s)|Mg²⁺(aq)║Sn²⁺(aq)|Sn(s)
Therefore, the cell notation for the voltaic cell with the given reaction is Mg(s)|Mg²⁺(aq)║Sn²⁺(aq)|Sn(s).
Learn more about voltaic cells here:
I hope it helps you!
Answer:
The element carbon is a part of seawater, the atmosphere, rocks such as limestone and coal, soils, as well as all living things. ... When plants and animals die, their bodies, wood and leaves decays bringing the carbon into the ground. Some is buried and will become fossil fuels in millions and millions of years.
Explanation:
The gas would diffuse as much as it could
Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka =
Replacing:
1,8x10⁻⁵ =
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!