Question

Pairs of electrodes and their corresponding standard electrode potentials are given. Arrange the cells constructed by these electrodes in increasing order of emf values. (1) Electrode \(\rightarrow \mathrm{A}, \mathrm{E}^{\circ}=-2.92 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{B}, \mathrm{E}^{\circ}=-2.71 \mathrm{~V}\) (2) Electrode \(\rightarrow \mathrm{C}, \mathrm{E}^{\circ}=-0.760 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{D}, \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\) (3) Electrode \(\rightarrow \mathrm{B}, \mathrm{E}^{\circ}=-2.71 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{D}, \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\) (4) Electrode \(\rightarrow \mathrm{A}, \mathrm{E}^{\circ}=-2.92 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{C}, \mathrm{E}^{\circ}=-0.76 \mathrm{~V}\) (a) 2143 (b) 1243 (c) 2413 (d) 2431

Short answer

Question: Arrange the cells formed by given electrode pairs in increasing order of emf values. Electrode pairs and their standard potentials are: A (-2.92 V), B (-2.71 V), C (-0.76 V), and D (-0.44 V). Answer: (b) 1243

Step-by-step solution

Compute the emf of each cell

To compute the emf of each cell, subtract the standard electrode potential of the anode from that of the cathode as given by the formula: emf = \(E^{\circ}_{cathode} - E^{\circ}_{anode}\). (1) Cell 1: \(\text{emf}_1 = E_{B}^\circ - E_{A}^\circ = (-2.71) - (-2.92) = 0.21 \mathrm{~V}\) (2) Cell 2: \(\text{emf}_2 = E_{D}^\circ - E_{C}^\circ = (-0.44) - (-0.760) = 0.32 \mathrm{~V}\) (3) Cell 3: \(\text{emf}_3 = E_{D}^\circ - E_{B}^\circ = (-0.44) - (-2.71) = 2.27 \mathrm{~V}\) (4) Cell 4: \(\text{emf}_4 = E_{C}^\circ - E_{A}^\circ = (-0.76) - (-2.92) = 2.16 \mathrm{~V}\)

Arrange the cells in increasing order of emf values

To arrange the cells in increasing order of emf values, sort the emf values computed in Step 1: \(\text{emf}_1 = 0.21 \mathrm{~V}\) \(\text{emf}_2 = 0.32 \mathrm{~V}\) \(\text{emf}_4 = 2.16 \mathrm{~V}\) \(\text{emf}_3 = 2.27 \mathrm{~V}\) So, keeping this in mind: 1. Cell 1 2. Cell 2 4. Cell 4 3. Cell 3 Thus, the correct option is (b) 1243.

Key concepts

emf calculation

Calculating the electromotive force (emf) of a galvanic cell is key to predicting its ability to drive an electrical current. The emf is determined using the standard electrode potentials of the cathode and the anode. The formula to compute the emf is:

• emf = \(E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)

This means you subtract the anode's potential from the cathode's potential. Since the direction of electron flow defines the anode and cathode, it is crucial first to identify which electrode will undergo reduction (becoming the cathode) and which will undergo oxidation (becoming the anode). When you perform this calculation, be mindful of the signs associated with each standard electrode potential. A positive emf indicates that the cell can perform work as it allows a spontaneous electron flow from anode to cathode.
By following this formula, you can predict which chemical reaction will occur spontaneously and how the cell can be utilized for practical purposes based on its calculated voltage.

cell construction

The construction of a galvanic cell involves pairing two electrodes, each composed of a different material, each existing in contact with an electrolyte. Each electrode acts either as an anode where oxidation occurs or a cathode where reduction occurs. Key aspects of cell construction include:

• Identifying the electrodes: Electrode materials with known standard electrode potentials are chosen.
• Separating half-cells: Each electrode's oxidation or reduction reaction is contained in a half-cell.
• Utilizing a salt bridge: A salt bridge completes the circuit while allowing ion flow to neutralize charge buildup.

The cell's physical setup allows electrons to flow externally through a conductor from the anode to the cathode, thus generating electrical power. The salt bridge not only maintains electrical neutrality but prevents the solutions in different half-cells from mixing which could otherwise disrupt the designated reactions.

oxidation-reduction reactions

Oxidation-reduction reactions, often referred to as redox reactions, are chemical processes where electrons are transferred between substances. These reactions form the basis of electrochemical cells, including galvanic cells. Understanding the roles of oxidation and reduction is essential:

• Oxidation: Loss of electrons by a substance, increasing its oxidation state.
• Reduction: Gain of electrons by a substance, decreasing its oxidation state.

In a galvanic cell, the substance that undergoes oxidation occurs at the anode, while the substance that undergoes reduction occurs at the cathode. This electron exchange creates a flow from anode to cathode, harnessed to provide electrical energy. Identifying which species is oxidized and which is reduced is fundamental to deciphering the reaction's direction and its respective electrodes' assignment within a cell.

standard electrode potentials

Standard electrode potentials are intrinsic values used to predict the direction of a redox reaction within electrochemical cells. Measured under standard conditions (298 K, 1 atm, and 1 M concentration), these potentials help classify electrodes as strong oxidants or reductants:

• Cathode: Typically associated with the higher potential, indicating a predisposition to gain electrons (reduction).
• Anode: Typically associated with the lower potential, indicating a predisposition to lose electrons (oxidation).

These potentials allow chemists to assess if a reaction will occur spontaneously. By referencing a chart of standard electrode potentials, you can construct a galvanic cell and predict the resulting emf and its suitability for powering devices. The convention is to use the standard hydrogen electrode, assigned a potential of 0 V, as a reference for these measurements, ensuring consistency in comparing various electrode materials.