Answer :
The correct answer is option B. [tex]Ni\(^{2+}\)[/tex]
To determine which metal ion solution would deposit the reduced metal mass onto the cathode the fastest under identical current and concentration conditions, we need to consider the concept of electrochemical equivalent or molar mass and the valency of the metal ions. The key factor is the number of moles of electrons required to deposit one mole of the metal (Faraday's law of electrolysis).
Faraday's law of electrolysis states that the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
- The relationship can be expressed as:
m = [tex]\frac{Q \cdot M}{n \cdot F}[/tex]
where:
- m is the mass of the substance deposited.
- Q is the total electric charge passed.
- M is the molar mass of the metal.
- n is the number of moles of electrons required to reduce one mole of the metal ion.
- F is Faraday's constant (approximately 96,485 C/mol).
The rate at which a metal deposits depends inversely on the product of its molar mass M and the number of electrons n required to reduce each ion. We can compare the metals based on their molar masses and valencies:
- [tex]Sn\(^{2+}[/tex] (Tin):
- Molar mass of Sn = 118.71 g/mol
- Valency (n) = 2
- [tex]Ni\(^{2+}\)[/tex] (Nickel):
- Molar mass of Ni = 58.69 g/mol
- Valency (n) = 2
- [tex]Pb\(^{2+}\)[/tex] (Lead):
- Molar mass of Pb = 207.2 g/mol
- Valency (n) = 2
- [tex]Cd\(^{2+}\)[/tex] (Cadmium):
- Molar mass of Cd = 112.41 g/mol
- Valency (n) = 2
- [tex]Ba\(^{2+}\)[/tex] (Barium):
- Molar mass of Ba = 137.33 g/mol
- Valency (n) = 2
To find out which metal deposits the fastest, we should look for the metal with the lowest [tex]\frac{M}{n}[/tex] ratio because a lower ratio indicates that less charge is required to deposit a given mass of the metal.
Calculate [tex]\frac{M}{n}[/tex] for each:
- [tex]Sn\(^{2+}[/tex]:
[tex]\( \frac{118.71}{2}[/tex] = 59.355
- [tex]Ni\(^{2+}\)[/tex]:
[tex]\( \frac{58.69}{2}[/tex] = 29.345
- [tex]Pb\(^{2+}\)[/tex]:
[tex]\( \frac{207.2}{2}[/tex] = 103.6
- [tex]Cd\(^{2+}\)[/tex]:
[tex]\( \frac{112.41}{2}[/tex] = 56.205
- [tex]Ba\(^{2+}\)[/tex]:
[tex]\( \frac{137.33}{2}[/tex] = 68.665
Based on these calculations, [tex]Ni\(^{2+}\)[/tex] has the lowest [tex]\frac{M}{n}[/tex] ratio of 29.345, indicating that Nickel will deposit the fastest under identical current and concentration conditions.
Final answer:
The solution that would deposit reduced metal mass onto the cathode the fastest under identical current and concentration conditions is Cd²+ (cadmium ion).
Explanation:
The solution that would deposit reduced metal mass onto the cathode the fastest under identical current and concentration conditions is D. cd2, which refers to cadmium ion (Cd²+). In an electrochemical cell, the reduction reaction occurs at the cathode, where the metal ion gains electrons and is deposited as a solid metal. The concentration of the metal ion in the half-cell solution also affects the rate at which the metal mass is deposited. Therefore, the higher the concentration of Cd²+ ions, the faster the reduced metal mass will be deposited onto the cathode.
