Daniell_Galvanic_Voltaic Cell 👉 Salt Bridge 👉 Equilibrium constant K - Reaction quotient Q +EXERCISE

How a Daniell Cell Works

A Daniell-type galvanic (voltaic) cell separates a spontaneous redox reaction into two half-cells, converting chemical energy into electrical energy.

In a zinc–copper cell:

• At the anode, zinc is oxidized: zinc atoms enter the solution as zinc ions and release electrons, increasing the zinc-ion concentration.
• At the cathode, copper ions are reduced: they gain electrons and become copper metal, decreasing the copper-ion concentration.
• Electrons travel through the external circuit from the zinc electrode to the copper electrode. With the voltmeter’s common terminal connected to zinc, the described setup reads about +1.1 V.
• Separating the zinc metal from the copper-ion oxidizing agent encourages electrons to travel through the external circuit rather than transfer through a local reaction.

Why the Salt Bridge Is Needed

The salt bridge, shown containing potassium chloride, allows ions to move between the solutions and maintains electrical neutrality as the reaction proceeds:

• Potassium cations migrate toward the cathode compartment, where copper ions are being removed from solution.
• Chloride anions migrate toward the anode compartment, where positively charged zinc ions are accumulating.

The general rule is that anions move toward the anode and cations move toward the cathode. Whether this appears clockwise or counterclockwise depends on how the cell is drawn.

The video notes that a porous membrane was considered, but a salt bridge proved more effective for connecting the two half-cells.

Exercise: Aluminum–Copper Cell

The exercise replaces zinc with aluminum and asks about cell polarity, electrode reactions, reaction direction, and the amount of aluminum consumed.

• With the meter’s common terminal connected to aluminum and a positive reading, aluminum is the negative terminal and copper is the positive terminal.
• Aluminum is oxidized at the anode, while copper ions are reduced at the cathode.
• The balanced overall reaction is:

[ 2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s) ]

• Pure solids are omitted when calculating the reaction quotient. For this reaction:

[ Q = \frac{[\text{Al}^{3+}]^2}{[\text{Cu}^{2+}]^3} ]

The exercise gives an initial reaction quotient of (Q = 10). Comparing (Q) with the equilibrium constant (K):

• If (Q < K), the reaction proceeds forward.
• If (Q > K), the reaction proceeds in reverse.
• If (Q = K), the system is at equilibrium.

The exercise indicates that (Q) is below (K), so the reaction proceeds forward, consistent with the identified cell polarity.

Charge and Electrode Mass

The video relates electric charge, electron transfer, and aluminum consumption:

• Charge transferred: (q = I \times \Delta t)
• Moles of electrons transferred: (n(e^-) = q/F), where (F) is the Faraday constant—the charge per mole of electrons.
• Aluminum oxidation: (\text{Al} \rightarrow \text{Al}^{3+} + 3e^-)
• Moles of aluminum consumed: (n(\text{Al}) = n(e^-)/3)

For an operating time of 1 hour 30 minutes at 40 mA, these relationships give approximately:

• 216 C of charge
• 0.00224 mol of electrons
• 0.020 g of aluminum consumed, using a molar mass of 27 g/mol

The same charge relationships can be used to discuss battery capacity and operating duration.

Speakers and Sources

• Speaker: An unnamed presenter/narrator from the PCCL channel
• Source/channel: PCCL – Physics – Chemistry
• Historical figure mentioned, but not heard speaking: Mr. Daniell