Question

The same quantity of electricity was passed through two separate electrolytic cells. The first of these contained a solution of copper sulfate \(\left(\mathrm{CuSO}_{4}\right)\) and exhibited the following cathode reaction (reduction): \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) The second of these contained a solution of silver nitrate \(\left(\mathrm{AgNO}_{3}\right)\) and exhibited the following cathode reaction: \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})\) If \(3.18 \mathrm{~g}\) of Cu were deposited in the first cell, how much (Ag) was deposited in the second cell?

Short answer

The amount of deposited silver in the second cell is \(10.79 g\).

Step-by-step solution

1. Calculate mole of deposited copper

We are given the mass of deposited copper \(3.18 g\). Using the molar mass of copper \(\left(\mathrm{M_{Cu}}=63.55 \frac{g}{mol}\right)\), we can calculate the number of moles of deposited copper. \[n_{\mathrm{Cu}}=\frac{\text { mass of deposited copper }}{\mathrm{M_{Cu}}}\]

2. Calculate total mole of electrons transferred in both cells

Using the given half-reaction where one mole of copper ion forms one mole of solid copper by using two moles of electrons, \[\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu}(\mathrm{s})\] Now, we can calculate the number of moles of electrons transferred in both cells as follows: \[n_{e} = 2 \cdot n_{\mathrm{Cu}}\]

3. Calculate mole of deposited silver

In the second cell, the half-reaction shows that one mole of silver ion forms one mole of solid silver by using one mole of electron: \[\mathrm{Ag}^{+}+\mathrm{e}^- \rightarrow \mathrm{Ag}(\mathrm{s})\] Using the discovered number of moles of electrons, \(n_{e}\), we can now calculate the number of moles of deposited silver: \[n_{\mathrm{Ag}} = n_{e}\]

4. Calculate mass of deposited silver

Finally, using the molar mass of silver \(\left(\mathrm{M_{Ag}}=107.87 \frac{g}{mol}\right)\), we can find the mass of deposited silver (\(m_{\mathrm{Ag}}\)): \[m_{\mathrm{Ag}} = n_{\mathrm{Ag}} \cdot \mathrm{M_{Ag}}\] Now, let's perform the calculations: Step 1: \[n_{\mathrm{Cu}} = \frac{3.18 g}{63.55 \frac{g}{mol}}= 0.050 mol\] Step 2: \[n_{e} = 2 \times 0.050 mol = 0.1 mol\] Step 3: \[n_{\mathrm{Ag}} = n_{e} = 0.1 mol\] Step 4: \[m_{\mathrm{Ag}} = n_{\mathrm{Ag}} \times 107.87 \frac{g}{mol} = 0.1 mol \times 107.87 \frac{g}{mol} = 10.79 g\] Therefore, the amount of deposited silver in the second cell is \(10.79 g\).

Key concepts

Mole of Deposited Copper

Understanding the concept of moles is vital in chemistry, especially when dealing with electrochemical reactions like the deposition of copper in an electrolytic cell. The 'mole' represents a specific quantity of anything, specifically Avogadro's number of particles, atoms, ions, or electrons, which is approximately 6.022 × 10²³.

When an electrolytic cell deposits copper, the reduction reaction that occurs at the cathode can be quantified in moles. To calculate the moles of copper deposited, one can use the formula:
\[n_{\mathrm{Cu}} = \frac{\text{mass of deposited copper}}{\mathrm{M_{Cu}}}\]
where \(n_{\mathrm{Cu}}\) is the number of moles and \(\mathrm{M_{Cu}}\) is the molar mass of copper. With the mass of the deposited copper known, this equation efficiently converts that value into moles, providing a way to link the mass with the number of individual copper atoms that make up that mass.

Total Mole of Electrons Transferred

When an electrolytic process occurs, such as the deposition of metals, a crucial factor to consider is the number of electrons that transfer during the reaction. This number is directly related to the stoichiometry of the chemical reaction at the cathode.

In the given exercise, the electrochemical reaction for copper deposition requires two moles of electrons for every one mole of copper ions that is reduced to solid copper. This is described by the chemical equation:
\[\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu}(\mathrm{s})\]
Using the equation
\[n_{e} = 2 \cdot n_{\mathrm{Cu}}\]
we quantify the total moles of electrons transferred during the reaction. This step is critical because the same quantity of electricity implies the same moles of electrons will be used in the silver deposition, allowing us to connect the processes in the two separate electrolytic cells.

Mole of Deposited Silver

Similar to copper in the first cell, the deposition of silver in the second cell also hinges on the concept of moles. However, for silver, the stoichiometry differs - only one mole of electrons is needed to reduce one mole of silver ions to solid silver, as shown by the reaction:
\[\mathrm{Ag}^{+}+\mathrm{e}^- \rightarrow \mathrm{Ag}(\mathrm{s})\]
Therefore, the moles of silver deposited, \(n_{\mathrm{Ag}}\), can be found by simply taking the total moles of electrons transferred to the silver cell, \(n_{e}\), which is a direct one-to-one conversion. Because the quantity of electricity is conserved across both cells, understanding this concept allows us to directly relate the amount of copper deposited to the resulting amount of silver deposited.

Mass of Deposited Metal

Once the amount of metal deposited is known in moles, it becomes necessary to convert this value into a more tangible quantity – the mass. The mass of a deposited metal can be calculated by using the molar mass of the metal and the number of moles of deposited metal.

The general equation for this conversion is:
\[m_{\mathrm{metal}} = n_{\mathrm{metal}} \cdot \mathrm{M_{metal}}\]
where \(m_{\mathrm{metal}}\) represents the mass of the deposited metal, \(n_{\mathrm{metal}}\) is the number of moles, and \(\mathrm{M_{metal}}\) is the molar mass of the metal. By applying this equation to the silver deposition reaction, we're able to determine the exact mass of silver that was deposited in the second cell, concluding the set of calculations required to bridge the gap between moles, electrons, and mass in electrolytic cell processes.