Question

chromium (iii) or chromium (vi) are frequently used to apply chrome finish to sink fixtures such as faucets. if 45.2 amps flows through a solution of chromium (iii) for 2 hours, how many grams of chromium can be deposited on a fixture?
a) 58.45 g
b) 0.974 g
c) 175.35 g
d) 0.016 g

Explanation:

Step1: Calculate total charge (Q)

Current \( I = 45.2 \, \text{A} \), time \( t = 2 \, \text{hours} = 2 \times 3600 \, \text{s} = 7200 \, \text{s} \).
Using \( Q = I \times t \),
\( Q = 45.2 \, \text{A} \times 7200 \, \text{s} = 325440 \, \text{C} \).

Step2: Determine moles of electrons

Faraday's constant \( F = 96485 \, \text{C/mol e}^- \).
Moles of \( \text{e}^- = \frac{Q}{F} = \frac{325440 \, \text{C}}{96485 \, \text{C/mol e}^-} \approx 3.373 \, \text{mol e}^- \).

Step3: Relate electrons to Cr(III) reduction

Reduction of \( \text{Cr}^{3+} \): \( \text{Cr}^{3+} + 3\text{e}^-
ightarrow \text{Cr} \).
Moles of \( \text{Cr} = \frac{\text{moles of e}^-}{3} = \frac{3.373 \, \text{mol e}^-}{3} \approx 1.124 \, \text{mol Cr} \).

Step4: Calculate mass of Cr

Molar mass of \( \text{Cr} = 52.00 \, \text{g/mol} \).
Mass \( = \text{moles} \times \text{molar mass} = 1.124 \, \text{mol} \times 52.00 \, \text{g/mol} \approx 58.45 \, \text{g} \).

Answer:

A) 58.45 g