Question

how many seconds will it take to reduce 1.0 × 10³ g al from an aqueous al⁺³ solution at a current of 100.0 a?
al³⁺ + 3e⁻ → al
e_red = −1.66 v
t = ? sec

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is 26.98 g/mol. Moles of Al, \( n = \frac{mass}{molar\ mass} = \frac{1.0\times10^{3}\ g}{26.98\ g/mol} \approx 37.06\ mol \)

Step2: Determine moles of electrons

From the reaction \( \text{Al}^{3+} + 3e^- ightarrow \text{Al} \), 3 moles of electrons are needed per mole of Al. So moles of electrons, \( n_{e^-} = 3\times n_{\text{Al}} = 3\times37.06\ mol = 111.18\ mol \)

Step3: Calculate charge (in coulombs)

Using Faraday's constant \( F = 96485\ C/mol \), charge \( Q = n_{e^-} \times F = 111.18\ mol\times96485\ C/mol \approx 1.072\times10^{7}\ C \)

Step4: Calculate time

Current \( I = 100.0\ A = 100.0\ C/s \). Time \( t = \frac{Q}{I} = \frac{1.072\times10^{7}\ C}{100.0\ C/s} = 1.072\times10^{5}\ s \) (approximate, more precise calculation below)

More precise calculation:
Moles of Al: \( \frac{1000\ g}{26.9815\ g/mol} \approx 37.062\ mol \)
Moles of electrons: \( 37.062\ mol\times3 = 111.186\ mol \)
Charge: \( 111.186\ mol\times96485\ C/mol = 111.186\times96485\ C \approx 10720000\ C \) (exact: \( 111.186\times96485 = 111.186\times(90000 + 6000 + 485)= 111.186\times90000 + 111.186\times6000 + 111.186\times485 = 9996740 + 667116 + 53925.21 = 10717781.21\ C \))
Time: \( \frac{10717781.21\ C}{100\ C/s} = 107177.8121\ s \approx 1.07\times10^{5}\ s \) (or more accurately, using exact values:

\( n_{\text{Al}} = \frac{1000}{26.981538} \approx 37.06209\ mol \)

\( n_{e^-} = 37.06209\times3 = 111.18627\ mol \)

\( Q = 111.18627\times96485.33212 = 111.18627\times96485.33212 \)

Calculate that: 96485.33212 * 100 = 9,648,533.212; 96485.33212 * 11.18627 ≈ 96485.33212*10=964,853.3212; 96485.33212*1.18627≈96485.33212*1=96,485.33212; 96485.33212*0.18627≈18,070. So total ≈ 964,853.3212 + 96,485.33212 + 18,070 ≈ 1,079,408.65. So total Q ≈ 9,648,533.212 + 1,079,408.65 ≈ 10,727,941.86 C

Then t = 10,727,941.86 / 100 = 107,279.4186 s ≈ 1.07×10⁵ s (or more precisely, ~1.07×10⁵ s, or using significant figures: the given values are 1.0×10³ g (2 sig figs), 100.0 A (4 sig figs), molar mass of Al is known precisely, Faraday's constant is known precisely. So the limiting factor is 1.0×10³ g (2 sig figs), so the answer should have 2 sig figs? Wait, 1.0×10³ has two sig figs, 100.0 has four. So maybe two or three? Wait, 1.0×10³ is two, so the answer should be two sig figs: 1.1×10⁵? Wait no, let's check the calculation again.

Wait, 1.0×10³ g is two sig figs, 100.0 A is four. Molar mass of Al is 26.98 g/mol (four sig figs). So when calculating moles: 1000 g / 26.98 g/mol = 37.06 mol (four sig figs from molar mass, two from mass? Wait, 1.0×10³ is two sig figs, so 1000 g is two sig figs (the.0 is significant). So 1.0×10³ has two sig figs, so moles of Al is 37 mol (two sig figs? Wait, 1.0×10³ is two sig figs, so 1000 g is 1.0×10³, so two sig figs. Then moles of Al is (1.0×10³ g) / (27 g/mol) ≈ 37 mol (two sig figs? Wait, 26.98 is ~27, so 1000/27 ≈ 37.03, but with two sig figs, 37 mol (or 3.7×10¹). Then moles of electrons is 3×37 = 111 mol (but with two sig figs, 1.1×10² mol). Then charge is 1.1×10² mol × 96485 C/mol ≈ 1.06×10⁷ C. Then time is 1.06×10⁷ C / 100 A = 1.06×10⁵ s, which rounds to 1.1×10⁵ s? Wait, but maybe the problem expects using more precise values. Let's do it with exact steps:

1. Moles of Al: \( n = \frac{mass}{molar\ mass} = \frac{1.0\times10^3\ g}{26.98\ g/mol} \approx 37.06\ mol \) (using 26.98 for molar mass)

2. Moles of electrons: Each Al³⁺ gains 3 electrons, so \( n_{e^-} = 3 \times n_{\text{Al}} = 3 \times 37.06 = 111.18\ mol \)

3. Charge (Q) is given by \( Q = n_{e^-} \times F \), where \( F = 96485\ C/mol \) (Faraday's constant)

So \( Q = 111.18\ mol \times 96485\ C/mol = 111.18 \times 96485\ C \)

Calculate that: 111.18 * 96485 = (100 + 11.18) * 96485 = 100*96485 + 11.18*96485 = 9,648,500 + (11*96485 + 0.18*96485) = 9,648,500 + (1,061,335 + 17,367.3) = 9,648,500 + 1,078,702.3 = 10,727,202.3\ C

4. Time (t) is \( t = \frac{Q}{I} \), where \( I = 100.0\ A = 100.0\ C/s \)

So \( t = \frac{10,727,202.3\ C}{100.0\ C/s} = 107,272.023\ s \approx 1.07\times10^5\ s \) (or 107,000 s when rounded to three significant figures, since 1.0×10³ has two, 100.0 has four, so the least number of significant figures is two? Wait, 1.0×10³ is two sig figs, 100.0 is four, molar mass is ~27 (two sig figs if we take 27, but 26.98 is four). So the limiting factor is 1.0×10³ (two sig figs), so the answer should have two sig figs: 1.1×10⁵ s? Wait, no, 1.0×10³ is two sig figs, 100.0 is four, so when multiplying/dividing, the result should have two sig figs. Wait, 1.0×10³ is two, 26.98 is four, so 1.0×10³ / 26.98 is two sig figs: 37 (two sig figs: 3.7×10¹). Then 3.7×10¹ * 3 = 1.1×10² (two sig figs). Then 1.1×10² * 96485 = 1.06×10⁷ (two sig figs: 1.1×10⁷). Then time is 1.1×10⁷ / 100 = 1.1×10⁵ s. But when we calculated with more precise numbers, we got ~107,272 s, which is ~1.07×10⁵ s, which is three sig figs. Since 1.0×10³ has two, 100.0 has four, maybe the problem expects three sig figs because 1.0×10³ is two (but the.0 is significant, so two), 100.0 is four, molar mass is four. So perhaps the answer is approximately 1.07×10⁵ s, or 107,000 s.

But let's do the calculation with exact molar mass of Al (26.981538 g/mol) and exact Faraday's constant (96485.33212 C/mol):

Moles of Al: 1000 g / 26.981538 g/mol = 37.06209 mol

Moles of electrons: 37.06209 * 3 = 111.18627 mol

Charge: 111.18627 mol * 96485.33212 C/mol = 111.18627 * 96485.33212 C

Calculate that: 96485.33212 * 111.18627 = let's compute 96485.33212 * 100 = 9,648,533.212; 96485.33212 * 11.18627 = 96485.33212*(10 + 1.18627) = 964,853.3212 + 96485.33212*1.18627 ≈ 964,853.3212 + 114,431.99 ≈ 1,079,285.31. So total charge is 9,648,533.212 + 1,079,285.31 ≈ 10,727,818.52 C

Time: 10,727,818.52 C / 100.0 A = 107,278.1852 s ≈ 1.07×10⁵ s (or 107,000 s when rounded to three significant figures)

So the time is approximately 1.07×10⁵ seconds, or 107,000 seconds.

Answer:

\boxed{1.07\times10^{5}} (or \boxed{107000} depending on significant figures, but more accurately, using the calculation above, it's approximately 1.07×10⁵ seconds)