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The mass of silver to be deposited is 0.150 g - VCE - SSCE Chemistry - Question 10 - 2005 - Paper 1

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The mass of silver to be deposited is 0.150 g. If the current is held steady at 1.50 amps, the time, in seconds, that it takes to complete the plating is closest to ... show full transcript

Worked Solution & Example Answer:The mass of silver to be deposited is 0.150 g - VCE - SSCE Chemistry - Question 10 - 2005 - Paper 1

Step 1

Calculate the number of moles of silver to be deposited

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Answer

First, we need to convert the mass of silver to moles. The molar mass of silver (Ag) is approximately 107.87 g/mol. Therefore, the number of moles (n) is calculated as:

n=massmolar mass=0.150 g107.87 g/mol≈0.00139 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{0.150 \text{ g}}{107.87 \text{ g/mol}} \approx 0.00139 \text{ mol}

Step 2

Calculate the total charge needed for deposition

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Answer

The charge (Q) required to deposit silver can be calculated using Faraday's law, given that each mole of silver requires 1 Faraday (approximately 96485 C). Thus:

Q=n×F=0.00139 mol×96485 C/mol≈134extCQ = n \times F = 0.00139 \text{ mol} \times 96485 \text{ C/mol} \approx 134 ext{ C}

Step 3

Calculate the time required using the formula Q = It

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Answer

From the formula, we can find time:

t=QI=134 C1.50 A≈89.33 secondst = \frac{Q}{I} = \frac{134 \text{ C}}{1.50 \text{ A}} \approx 89.33 \text{ seconds}

Rounding this to the closest option given in the question, the answer is approximately 90 seconds, which corresponds to option A.

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