Q21.113CP - Chemistry: Molecular Nature Of Matter And Change

Question

An inexpensive and accurate method of measuring the quantity of electricity passing through a circuit is to pass it through a solution of a metal ion and weigh the metal deposited. A silver electrode immersed in an ${\mathbf{\text{Ag}}}^{\mathbf{\text{ + }}}$solution weighs $\mathbf{1}\mathbf{.}\mathbf{7854}\mathbf{ }\mathbf{g}$ before the current passes and weighs $\mathbf{1}\mathbf{.}\mathbf{8016}\mathbf{ }\mathbf{g}$ after the current has passed. How many coulombs have passed?

Step-by-Step Solution

Answer

The number of coulombs that have passed is 14.49 C.

Step 1: Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

An electrode is an electrical conductor that makes contact with the circuit's non-metallic components, such as an electrolyte, semiconductor, or vacuum. This is also known as an anode or cathode in an electrochemical cell.

Step 2: Information Provided

• The weight of Ag electrode, before the current is 1.7854 g.
• The weight after the current is 1.8016 g.
• Faraday constant: charge of 1 mole of electrons ($\text{F = 96485 C/mole}$)

Step 3: Calculation for Charge

The reaction given is –

${\text{Ag}}^{\text{ + }}{\text{(aq) + e}}^{\text{ - }}\rightleftharpoons \text{Ag(s)}$

The mass of Ag deposited is –

$\begin{array}{rcl}{\mathrm{m}}_{\mathrm{Ag}}& =& 1.8016 \mathrm{g}-1.7854\mathrm{g}\\ & =& 0.0162 \mathrm{g}\end{array}$

The number of moles of Ag deposited is (molar mass of Ag is 107.8682 g/mol) –

$\begin{array}{rcl}{\mathrm{n}}_{\mathrm{Ag}}& =& \frac{0.0162\mathrm{g}}{107.8682 \mathrm{g}/\mathrm{mol}}\\ & =& 1.502·{10}^{4 }\mathrm{mol} \mathrm{Ag}\end{array}$

Since 1 mole of electrons produces 1 mole of Ag, the number of moles of electrons is $1.502·{10}^{-4} \mathrm{mole}$.

Hence, calculate the charge using Faraday constant –

$\begin{array}{rcl}\mathrm{Charge}& =& \mathrm{Moles} {\mathrm{e}}^{-}·\mathrm{F}\\ & =& 1.502·{10}^{-4}\mathrm{mol} {\mathrm{e}}^{-}\\ & =& 14.49\mathrm{C}\end{array}$

 

Therefore, the value for charge is obtained as 14.49 C.