The mass of silver (Molar mass of $\mathrm{Ag}: 108 \mathrm{~gmol}^{-1}$ ) displaced by a quantity of electricity which displaces $5600 \mathrm{~mL}$ of $\mathrm{O}_2$ at S.T.P. will be ______ g.
Answer (integer)
108
Solution
<p>First, we need to determine the amount of $\mathrm{O}_2$ (oxygen gas) in moles that is displaced by the quantity of electricity mentioned. To do this, we'll use the molar volume of a gas at Standard Temperature and Pressure (S.T.P.), which is approximately $22.4 \mathrm{~L/mol}$ (or $22400 \mathrm{~mL/mol}$).</p>
<p>The volume of $\mathrm{O}_2$ is given as $5600 \mathrm{~mL}$. Now, we convert this volume to moles:</p>
$$
\text{moles of } \mathrm{O}_2 = \frac{\text{volume of } \mathrm{O}_2}{\text{molar volume}} = \frac{5600 \mathrm{~mL}}{22400 \mathrm{~mL/mol}} = 0.25 \mathrm{~mol}
$$
<p>Next, we'll use Faraday's laws of electrolysis to relate the moles of $\mathrm{O}_2$ to the moles of silver being displaced. In electrolysis, silver is deposited at the cathode according to the following half-reaction:</p>
$\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}$
<p>This tells us that for each mole of $\mathrm{Ag}^+$ ions, only 1 mole of electrons is required to reduce it to silver metal $\mathrm{Ag}$. However, the liberation of oxygen gas involves the following half-reaction:</p>
$2\mathrm{H_2O} (l) \rightarrow \mathrm{O}_2 (g) + 4H^+ (aq) + 4e^-$
<p>From this reaction, we can see that 1 mole of $\mathrm{O}_2$ gas requires 4 moles of electrons to be produced. Therefore, the number of moles of electrons associated with the $0.25 \mathrm{~mol}$ of $\mathrm{O}_2$ will be:</p>
$$
\text{moles of electrons for } \mathrm{O}_2 = 0.25 \mathrm{~mol} \times 4 = 1 \mathrm{~mol}
$$
<p>Now, since it takes 1 mole of electrons to reduce 1 mole of $\mathrm{Ag}^+$. The moles of electrons required is equal to the moles of $\mathrm{Ag}$ produced. Hence, we have 1 mole of $\mathrm{Ag}$ being deposited.</p>
<p>Finally, to calculate the mass of this silver, we use the molar mass of silver:</p>
$$
\text{Mass of Ag} = (\text{moles of Ag}) \times (\text{Molar mass of Ag}) = 1 \mathrm{~mol} \times 108 \mathrm{~gmol}^{-1} = 108 \mathrm{~g}
$$
<p>So, the mass of silver displaced by the quantity of electricity that displaces 5600 mL of $\mathrm{O}_2$ at S.T.P. will be $108 \mathrm{~g}$.</p>
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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