tal charge is:

$$ Q = I \times t $$

where:

$ I = 0.8 \, \text{A} $ (current)

$ t = 60 \times 60 = 3600 \, \text{s} $ (time converted to seconds)

So,

$$ Q = 0.8 \, \text{A} \times 3600 \, \text{s} = 2880 \, \text{Coulombs} $$

Step 2: Use Faraday's Second Law of Electrolysis.

The mass of substance deposited/liberated is given by:

$$ m = \frac{Q \times M}{n \times F} $$

where:

$ m $ = mass of substance deposited (in grams)

$ Q = 2880 \, \text{C} $ (total charge)

$ M = 40 \, \text{g/mol} $ (molar mass of calcium)

$ n = 2 $ (number of electrons exchanged; calcium forms $ \text{Ca}^{2+} $)

$ F = 96500 \, \text{C/mol} $ (Faraday's constant)

Step 3: Calculate the mass of calcium deposited.

Substituting in the values:

$$ m = \frac{2880 \times 40}{2 \times 96500} $$

Simplifying:

$$ m = \frac{115200}{193000} $$

$$ m \approx 0.5969 \, \text{g} $$

Rounding to one decimal place gives:

$$ m \approx 0.6 \, \text{g} $$

Therefore, the mass of calcium deposited at the cathode is approximately 0.6 grams.

Option C: 0.6 g

Electrochemistry

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