Question

A time variable current is passed through an aqueous solution of $AgNO_3$ for 60 s. The dependence is i= 125-t(s).

find the weight of silver deposited in the interval of 20 s - 40 s after the start of electrolysis. (Ag = 108).

Answer 1

2.1264 g

Answer 2

1. Understand the given information:

• Current, $i = 125 - t$ (Amperes), where $t$ is in seconds.
• Electrolysis of aqueous $AgNO_3$.
• Time interval for deposition: $t_1 = 20$ s to $t_2 = 40$ s.
• Atomic weight of Silver (Ag) = 108 g/mol.
• Faraday's constant, $F \approx 96500$ C/mol.

2. Determine the electrochemical reaction and equivalent weight of Silver:

At the cathode, silver ions ($Ag^+$) are reduced to silver metal ($Ag$):

$Ag^+(aq) + e^- \rightarrow Ag(s)$

The n-factor (number of electrons involved per mole of substance) for silver is 1.

The equivalent weight ($E$) of silver is its molar mass divided by its n-factor:

$E_{Ag} = \frac{\text{Molar Mass of Ag}}{\text{n-factor}} = \frac{108 \text{ g/mol}}{1} = 108 \text{ g/eq}$

3. Calculate the total charge (Q) passed during the given time interval:

Since the current is time-variable, the total charge $Q$ is found by integrating the current over the time interval:

$Q = \int_{t_1}^{t_2} i(t) \, dt$

$Q = \int_{20}^{40} (125 - t) \, dt$

Integrate the expression:

$Q = \left[ 125t - \frac{t^2}{2} \right]_{20}^{40}$

Now, substitute the upper and lower limits:

$Q = \left( 125 \times 40 - \frac{40^2}{2} \right) - \left( 125 \times 20 - \frac{20^2}{2} \right)$

$Q = \left( 5000 - \frac{1600}{2} \right) - \left( 2500 - \frac{400}{2} \right)$

$Q = (5000 - 800) - (2500 - 200)$

$Q = 4200 - 2300$

$Q = 1900 \text{ C}$

4. Calculate the weight of silver (W) deposited using Faraday's First Law of Electrolysis:

Faraday's First Law states:

$W = \frac{E \cdot Q}{F}$

Where:

• $W$ = weight of substance deposited
• $E$ = equivalent weight of the substance
• $Q$ = total charge passed
• $F$ = Faraday's constant ($96500 \text{ C/mol}$)

Substitute the calculated values:

$W = \frac{108 \text{ g/eq} \times 1900 \text{ C}}{96500 \text{ C/eq}}$

$W = \frac{108 \times 1900}{96500}$

$W = \frac{108 \times 19}{965}$

$W = \frac{2052}{965}$

$W \approx 2.1264 \text{ g}$

The weight of silver deposited is approximately 2.1264 g.