By what mass (in grams) will a chromium cathode increase when it is coupled to a magnesium half cell in which the magnesium anode loses 1.53 grams? (assuming chromium ions in solution are Cr^3+).
96,500 C will deposit 24.3/2 = 12.15 g Mg metal.
96, 500 C will deposit 52/3 = 17.33 g Cr
1.53 g Mg loss x (17.33/12.15) = 2.182 is the shortcut way to solve this. I can show you the long way if you want to see it. Actually, the long way is easier to undestand, I think.
I would greatly appreciate it if you could show me the long way walkthrough. Thank you!
To determine the mass gained by the chromium cathode when coupled to a magnesium half cell, we need to use Faraday's laws of electrolysis.
1. Determine the balanced half-reaction for the cathode:
The magnesium anode loses electrons and forms magnesium ions. The balanced half-reaction for the anode is:
Mg(s) → Mg^2+(aq) + 2e^-
Therefore, the balanced half-reaction for the cathode should involve the reduction of chromium(III) ions (Cr^3+) to chromium metal (Cr). Since chromium(III) ions gain electrons, the balanced half-reaction for the cathode is:
3e^- + Cr^3+(aq) → Cr(s)
2. Calculate the number of moles of electrons transferred:
From the balanced cathode half-reaction, we can see that 3 moles of electrons are involved for every 1 mole of Cr^3+ reduced. Since 1 mole of electrons corresponds to 1 Faraday (F), we can say that 3 moles of electrons correspond to 3 Faradays (3F).
3. Convert the mass change of the anode into moles:
The mass change of the magnesium anode is given as 1.53 grams. To convert this mass into moles, we need to divide it by the molar mass of magnesium (24.31 g/mol):
Number of moles of Mg = 1.53 g / 24.31 g/mol
4. Determine the moles of Cr^3+ reduced:
Since 3 moles of electrons correspond to the reduction of 1 mole of Cr^3+, the number of moles of Cr^3+ reduced is the same as the number of moles of electrons transferred, which is 3 times the moles of Mg calculated in step 3.
Number of moles of Cr^3+ reduced = 3 * (1.53 g / 24.31 g/mol)
5. Calculate the mass gained by the chromium cathode:
Finally, we can calculate the mass gained by the chromium cathode using the number of moles of Cr^3+ reduced and the molar mass of chromium (52.00 g/mol):
Mass gained by the chromium cathode = (Number of moles of Cr^3+ reduced) * (Molar mass of Cr)
Mass gained by the chromium cathode = [3 * (1.53 g / 24.31 g/mol)] * 52.00 g/mol
Calculating this expression will give you the mass gained by the chromium cathode.
To find the mass of chromium that will increase, we need to determine the corresponding change in oxidation state for the chromium ions.
In the given question, we are told that the magnesium anode loses 1.53 grams. However, we need to convert this mass of magnesium to moles. To do this, we divide the mass by the molar mass of magnesium.
The molar mass of magnesium (Mg) is approximately 24.31 g/mol.
Therefore, the number of moles of magnesium can be calculated as follows:
moles of magnesium = mass of magnesium (g) / molar mass of magnesium (g/mol)
= 1.53 g / 24.31 g/mol
≈ 0.063 mol
Since the chromium cathode is coupled to the magnesium half cell, the number of electrons exchanged in the reaction is the same for both elements. In this case, the number of electrons transferred is 3, corresponding to Cr^3+ ions.
The number of moles of electrons (n) involved in the reaction can be calculated using Faraday's constant, which is approximately 96485 C/mol. Since each mole of electrons is equivalent to one Faraday of charge, we can determine the number of moles of electrons using the equation:
n = Faraday constant (C/mol) × charge (C)
= 96485 C/mol × 3
= 289455 C
Now, using Faraday's law, we can determine the mass of chromium deposited. Since every 3 moles of electrons correspond to 1 mole of Cr^3+ ions, the moles of chromium (Cr) will be one-third of the moles of electrons (n).
moles of chromium = n / 3
= 289455 C / 3
≈ 96485 mol
Finally, to find the mass of chromium deposited, we multiply the moles of chromium by the molar mass of chromium.
The molar mass of chromium (Cr) is approximately 51.996 g/mol.
mass of chromium = moles of chromium × molar mass of chromium
= 96485 mol × 51.996 g/mol
≈ 5020.69 g
Therefore, the chromium cathode will increase by approximately 5020.69 grams when coupled to the magnesium half cell.