Question

A 1 M solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A 1 \(\mathrm{M}\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass, and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Short answer

(a) The Cu electrode serves as the cathode, and the Sn electrode serves as the anode. (b) The Cu electrode gains mass, and the Sn electrode loses mass as the cell reaction proceeds. (c) The overall cell reaction is: Sn + Cu²⁺ → Sn²⁺ + Cu (d) The emf generated by the cell under standard conditions is \(0.48\,\text{V}\).

Step-by-step solution

(Step 1: Identify the half-reactions)

First, we need to identify the two half-reactions occurring at the electrodes. From the provided solutions, we have the following half-reactions: Cu²⁺ + 2e⁻ → Cu (Reduction half-reaction) Sn²⁺ + 2e⁻ → Sn (Reduction half-reaction)

(Step 2: Compare the reduction potentials)

We will consult a table of reduction potentials to determine which half-reaction would occur spontaneously. Here are the standard reduction potentials at 25°C: Cu²⁺ + 2e⁻ → Cu (Reduction): \(E^0_{red} = 0.34\) V Sn²⁺ + 2e⁻ → Sn (Reduction): \(E^0_{red} = -0.14\) V Since Cu²⁺ has a higher reduction potential, it would be more favorable to undergo the reduction and therefore Cu will act as the cathode. Conversely, Sn²⁺ will undergo the oxidation process, making it the anode.

(Step 3: Identify the overall reaction)

To find the overall reaction, we need to combine the reduction and oxidation half-reactions. We will balance the charges and add the reactions: Oxidation half-reaction: Sn → Sn²⁺ + 2e⁻ Reduction half-reaction: Cu²⁺ + 2e⁻ → Cu Overall cell reaction: Sn + Cu²⁺ → Sn²⁺ + Cu

(Step 4: Find the cell's emf )

The emf generated by a cell under standard conditions can be calculated using the difference of the standard reduction potentials: \(E^0_{cell} = E^0_{red(cathode)} - E^0_{red(anode)}\) In this case, \(E^0_{cell} = 0.34\,\text{V} - (-0.14\,\text{V}) = 0.48\,\text{V}\)

(Step 5: Identify the electrodes that gain and lose mass)

The cathode electrode gains mass as Cu²⁺ ions are reduced to metallic Cu and deposit on it. The anode electrode loses mass as the Sn metal is oxidized to Sn²⁺ ions, leaving the metal surface. (a) The Cu electrode serves as the cathode, and the Sn electrode serves as the anode. (b) The Cu electrode gains mass, and the Sn electrode loses mass as the cell reaction proceeds. (c) The overall cell reaction is: Sn + Cu²⁺ → Sn²⁺ + Cu (d) The emf generated by the cell under standard conditions is \(0.48\,\text{V}\).

Key concepts

Electrode Reactions

In a galvanic cell, chemical reactions occur at each electrode, and these are known as half-reactions. In the beaker with copper, copper ions (\(\text{Cu}^{2+}\)) in the solution gain electrons to form solid copper. This is a reduction half-reaction because the copper ions gain electrons:

• \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

The other beaker contains tin, where solid tin loses electrons to form tin ions (\(\text{Sn}^{2+}\)) in the solution. This is called an oxidation half-reaction:

• \[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \]

• The electrode where oxidation occurs is called the anode.
• The electrode where reduction occurs is called the cathode.

In our exercise, the tin electrode acts as the anode because it undergoes oxidation, while the copper electrode acts as the cathode as it undergoes reduction.

Standard Reduction Potentials

Standard reduction potentials help us determine how readily a species will gain electrons. Each half-reaction has a standard reduction potential, denoted as \(E^0_{red}\), measured in volts. This value indicates a substance's tendency to be reduced. For copper and tin:

• Copper's reduction potential: \(E^0_{red(\text{Cu}^{2+/Cu})} = 0.34 \, \text{V}\)
• Tin's reduction potential: \(E^0_{red(\text{Sn}^{2+/Sn})} = -0.14 \, \text{V}\)

The higher the reduction potential, the more efficient the species is at gaining electrons. Between copper and tin, copper has a higher reduction potential, meaning it is more likely to get reduced compared to tin. Thus, the copper solution forms the cathode as it accepts electrons. Knowing these values helps us predict the flow of electrons in a galvanic cell, which is essential for determining the anode and cathode in electrochemical reactions.

Cell EMF

The electromotive force (emf) of a cell is the energy per unit charge available due to the chemical reaction. It is calculated as the difference between the standard reduction potentials of the cathode and the anode. The formula is:

• \[ E^0_{cell} = E^0_{red(cathode)} - E^0_{red(anode)} \]

For our galvanic cell:

• Cathode: Copper, \(E^0_{red} = 0.34 \, \text{V}\)
• Anode: Tin, \(E^0_{red} = -0.14 \, \text{V}\)

Using these values, we compute the cell emf:

• \[ E^0_{cell} = 0.34 \, \text{V} - (-0.14 \, \text{V}) = 0.48 \, \text{V} \]

This value represents the maximum potential difference between the electrodes, and it signals the spontaneity of the electrochemical reaction. A positive cell emf indicates that the reaction can proceed spontaneously, providing energy for electric current to flow through the external circuit.