Question

a current of 20.0 a is applied to a zn⁺² solution. how many seconds will it take to reduce 15.0 grams of zinc?
zn²⁺ + 2e⁻ → zn
e_red = −0.76 v
t = ? sec

Explanation:

Step 1: Calculate moles of Zn

Molar mass of Zn is 65.38 g/mol. Moles of Zn = mass / molar mass = \( \frac{15.0\ g}{65.38\ g/mol} \approx 0.2294\ mol \).

Step 2: Determine moles of electrons

From the reaction \( Zn^{2+} + 2e^- \to Zn \), 2 moles of electrons are needed per mole of Zn. So moles of \( e^- \) = \( 2 \times 0.2294\ mol = 0.4588\ mol \).

Step 3: Calculate charge (in coulombs)

Using Faraday's constant (\( F = 96485\ C/mol\ e^- \)), charge \( Q = n \times F \), where \( n \) is moles of electrons. So \( Q = 0.4588\ mol \times 96485\ C/mol \approx 44270\ C \).

Step 4: Find time using \( I = \frac{Q}{t} \)

Current \( I = 20.0\ A = 20.0\ C/s \). Rearranging \( t = \frac{Q}{I} \), so \( t = \frac{44270\ C}{20.0\ C/s} = 2213.5\ s \approx 2210\ s \) (or more precisely calculated).

Answer:

\( \approx 2210\) (or more accurately, using precise calculations: first, moles of Zn: \( \frac{15.0}{65.38} \approx 0.2294\), moles of \( e^- \): \( 0.2294\times2 = 0.4588\), charge: \( 0.4588\times96485 = 44267\ C\), time: \( \frac{44267}{20.0} = 2213.35\ s \approx 2210\ s \) or keeping more decimals as needed. The exact calculation gives approximately 2210 seconds (or 2213 seconds depending on precision).)