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Zorluk: Çok zorFaraday's Laws of Electrolysis and Quantitative Calculations

An aqueous solution of chromium(III) tetraoxosulfate(VI) is electrolyzed using inert platinum electrodes. A steady current of 5.00 A5.00\text{ A} is passed through the electrolyte for 96.5 minutes96.5\text{ minutes}. If the cathodic current efficiency for chromium deposition is 75.0%75.0\%, calculate the mass, in grams, of chromium metal deposited at the cathode. [Molar mass of Cr=52.0 g/mol\text{Cr} = 52.0\text{ g/mol}, 1 F=96500 C/mol1\text{ F} = 96500\text{ C/mol}]

Cevap: 3.9 g

Cevap

3.90 g3.90\text{ g}
To find the mass of chromium deposited, calculate total charge (Q=I×t=5.00×5790=28950 CQ = I \times t = 5.00 \times 5790 = 28950\text{ C}), adjust for 75.0%75.0\% current efficiency (Qeff=21712.5 CQ_{\text{eff}} = 21712.5\text{ C}), convert to Faradays (0.225 F0.225\text{ F}), divide by the valency of 3 for Cr3+\text{Cr}^{3+} to find moles of chromium (0.075 mol0.075\text{ mol}), and multiply by molar mass (52.0 g/mol52.0\text{ g/mol}) to yield 3.90 g3.90\text{ g}.

Adım Adım Çözüm

1
Convert the electrolysis time into seconds
t=96.5×60=5790 st = 96.5 \times 60 = 5790\text{ s}
Standard SI unit of time (seconds) is required for charge calculation (Q=I×tQ = I \times t).
2
Calculate the total charge transferred
Q=5.00 A×5790 s=28950 CQ = 5.00\text{ A} \times 5790\text{ s} = 28950\text{ C}
Determines total quantity of electricity delivered by the current source.
3
Apply the current efficiency percentage
Qeff=28950 C×0.750=21712.5 CQ_{\text{eff}} = 28950\text{ C} \times 0.750 = 21712.5\text{ C}
Only 75% of the total current is utilized specifically for reducing chromium ions.
4
Convert effective charge into moles of electrons
ne=21712.5 C96500 C/mol=0.225 mol e−n_e = \frac{21712.5\text{ C}}{96500\text{ C/mol}} = 0.225\text{ mol } e^-
Faraday's constant gives the charge carried per mole of electrons.
5
Relate moles of electrons to moles of chromium deposited
nCr=0.2253=0.075 moln_{\text{Cr}} = \frac{0.225}{3} = 0.075\text{ mol}
Reduction of one mole of Cr3+\text{Cr}^{3+} requires three moles of electrons (3 Faradays).
6
Calculate the mass of chromium deposited
m=0.075 mol×52.0 g/mol=3.90 gm = 0.075\text{ mol} \times 52.0\text{ g/mol} = 3.90\text{ g}
Mass is obtained by multiplying the number of moles by the molar mass.

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Faraday's Laws of Electrolysis and Current Efficiency
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Faraday's Laws of Electrolysis and Quantitative Calculations

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