An electrochemical cell is created using gold and magnesium half-cells.
a. Determine which half-cell will undergo oxidation and which will undergo reduction,
identify anode and cathode, and calculate the voltage for the cell. You do not need to
diagram the cell.
b. If the mass of the magnesium electrode changes by 5.0 g, what will be the change in
mass of the gold electrode, and will its mass increase or decrease?
A lot of typing with symbols etc. How can you do on your own?
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Well I'm trying to figure out the chemical equation
Mg|Mg^+2||Au^+3|Au
3Mg(s) + 2Au^+3 ==> 2Au(s) + 3Mg^+2
To determine which half-cell will undergo oxidation and which will undergo reduction, we need to look at the standard reduction potentials for the reactions involving gold and magnesium.
a. Oxidation occurs at the anode, and reduction occurs at the cathode. In an electrochemical cell, the species being oxidized loses electrons and becomes positively charged, while the species being reduced gains electrons and becomes negatively charged.
The standard reduction potential for gold (Au) is +1.50 V, while the standard reduction potential for magnesium (Mg) is -2.37 V.
Since gold has a higher reduction potential than magnesium, it means that gold is more likely to undergo reduction, and therefore, the gold half-cell will be the cathode.
Conversely, magnesium is more likely to undergo oxidation, so the magnesium half-cell will be the anode.
The voltage for the cell, also known as the cell potential (Ecell), can be determined by subtracting the reduction potential of the anode from the reduction potential of the cathode:
Ecell = E(cathode) - E(anode)
= +1.50 V - (-2.37 V)
= +1.50 V + 2.37 V
= 3.87 V
Hence, the voltage for the cell is 3.87 V.
b. To determine the change in mass of the gold electrode, we need to use Faraday's law of electrolysis, which states that the mass of a substance produced or consumed in an electrolytic reaction is directly proportional to the quantity of electricity passed through the system.
The equation for Faraday's law is given as:
Δm = (Z * F * ΔQ) / M
where:
Δm = change in mass
Z = number of moles of electrons transferred in the half-reaction
F = Faraday's constant (approximately 96,485 C/mol)
ΔQ = charge passed through the system (in coulombs)
M = molar mass of the substance
For magnesium, in its oxidation half-reaction, it loses two moles of electrons:
Mg(s) → Mg2+(aq) + 2e-
Therefore, Z = 2.
To calculate the change in mass, we need to know the charge passed through the system (ΔQ). We can find this by using Faraday's law and the given change in mass for magnesium.
Let's assume that the change in mass of magnesium is Δm = -5.0 g (negative because it is being consumed in the oxidation half-reaction).
ΔQ = (Δm * M * Z) / (F * 1)
Substituting the values:
ΔQ = (-5.0 g * 24.31 g/mol * 2) / (96,485 C/mol)
= -0.1027 C
Now, since gold is being reduced, it will gain electrons as follows:
Au3+(aq) + 3e- → Au(s)
The number of moles of electrons transferred (Z) for gold is 3.
To determine the change in mass of gold, we can rearrange the equation for Faraday's law:
Δm = (Z * F * ΔQ) / M
Substituting the known values:
Δm = (3 * 96,485 C/mol * -0.1027 C) / 196.97 g/mol (molar mass of gold)
= -0.150 g
Since the change in mass of gold is negative, it means that the mass of the gold electrode will decrease by 0.150 g.
Therefore, the change in mass of the gold electrode will be -0.150 g (decrease in mass).