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 \(1 M\) 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 the 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

The anode is the Sn electrode and the cathode is the Cu electrode. The Sn electrode loses mass, while the Cu electrode gains mass. The overall cell reaction is Sn(s) + Cu²⁺ → Sn²⁺ + Cu(s), and the emf generated under standard conditions is \( +0.48 V \).

Step-by-step solution

Determine the anode and cathode

In our cell, we have two half-cells. The first half-cell contains a solution of Cu(NO₃)₂ and Cu metal, while the second half-cell has a solution of SnSO₄ and Sn metal. To identify which electrode serves as the anode and which as the cathode, we will look at the standard electrode potentials of the two half-cells. For Cu²⁺ + 2e⁻ → Cu(s), the standard reduction potential E° is \( +0.34 V \) For Sn²⁺ + 2e⁻ → Sn(s), the standard reduction potential E° is \( -0.14 V \) Since the half-cell with a lower reduction potential will be oxidized (lose electrons) and serve as the anode, in this case, it's Sn(s) → Sn²⁺ + 2e⁻, and the anode is the Sn electrode. The Cu electrode will be the cathode.

Identify the electrode gaining and losing mass

The anode loses mass because the solid metal is oxidized and enters the solution. In our cell, the Sn metal loses mass because it undergoes the reaction: Sn(s) → Sn²⁺ + 2e⁻. At the cathode, Cu²⁺ ions in the solution gain electrons and are reduced to form solid Cu metal, causing the Cu electrode to gain mass. The reaction occurring at the cathode is: Cu²⁺ + 2e⁻ → Cu(s).

Write the overall cell reaction

To obtain the overall cell reaction, we will combine the oxidation and reduction half-cell reactions. Keep in mind that electrons should cancel out in the final cell reaction. Anode (oxidation): Sn(s) → Sn²⁺ + 2e⁻ Cathode (reduction): Cu²⁺ + 2e⁻ → Cu(s) Overall cell reaction: Sn(s) + Cu²⁺ → Sn²⁺ + Cu(s)

Calculate the emf generated by the cell under standard conditions

To calculate the emf generated by the cell under standard conditions, we can use the Nernst Equation for standard conditions: E°(cell) = E°(cathode) - E°(anode) We already know the standard reduction potentials for both half-cells: E°(Cu²⁺/Cu) = \( +0.34 V \) E°(Sn²⁺/Sn) = \( -0.14 V \) Now, we can input these values into the Nernst Equation: E°(cell) = (+0.34 V) - (-0.14 V) = +0.48 V The emf generated by the cell under standard conditions is \( +0.48 V \).

Key concepts

Redox Reactions

Redox reactions are fundamental to the functioning of electrochemical cells. In these reactions, one element is oxidized, while another is reduced. Oxidation refers to the loss of electrons, and reduction refers to the gain of electrons. A mnemonic to remember this is "OIL RIG": Oxidation Is Loss, Reduction Is Gain. In the given exercise, the redox reaction occurs between tin (Sn) and copper (Cu). While the tin loses electrons becoming oxidized to form Sn²⁺ ions, the copper ions gain electrons and are thus reduced to become solid copper metal. Understanding these electron transfers is key to figuring out how a galvanic cell functions.

Standard Electrode Potentials

Standard electrode potentials are values that indicate how readily a species gains electrons (is reduced) under standard conditions (1 M concentration, 1 atm pressure, and 25°C). They are measured in volts (V) and serve as a comparative tool to predict the direction of flow of electrons in a redox reaction. Every half-reaction has a standard electrode potential associated with it:

• For the reduction of Cu²⁺ to Cu, the potential is +0.34 V.
• For the reduction of Sn²⁺ to Sn, the potential is -0.14 V.

A more positive standard electrode potential means a greater tendency to be reduced. In our exercise, since Cu has a more positive value compared to Sn, it acts as the cathode, undergoing reduction, while Sn serves as the anode, undergoing oxidation.

Galvanic Cells

A galvanic cell, also known as a voltaic cell, is a device that converts chemical energy into electrical energy through a spontaneous redox reaction. It consists of two half-cells connected by a salt bridge, which facilitates the movement of ions and maintains electrical neutrality. In the exercise example, one half-cell has a copper electrode in a Cu(NO₃)₂ solution, and the other a tin electrode in a SnSO₄ solution.

• The anode is the Sn electrode, where oxidation occurs, resulting in a loss of mass as Sn atoms transform into Sn²⁺ ions.
• The cathode is the Cu electrode, where reduction occurs, gaining mass as Cu²⁺ ions turn into solid Cu.

These reactions generate an electric current, which can be captured for external work or measured using a voltmeter.

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry that calculates the electromotive force (emf) of a cell under non-standard conditions. However, when conditions are standard, the equation simplifies to this:\[ E^°(cell) = E^°(cathode) - E^°(anode) \]In the exercise scenario, it's underlined that the cell operates under standard conditions, making this equation particularly straightforward to use. By substituting the given standard potentials:

• For the cathode (Cu): +0.34 V
• For the anode (Sn): -0.14 V

You plug in these values to get the emf of:\[ E^°(cell) = (+0.34 \, V) - (-0.14 \, V) = +0.48 \, V \]This positive value firmly indicates that the cell reaction is spontaneous and capable of generating electrical energy effectively.