Chapter 19: Problem 73
A galvanic cell consists of a silver electrode in contact with \(346 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution and a magnesium electrode in contact with \(288 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. (a) Calculate \(E\) for the cell at \(25^{\circ} \mathrm{C}\). (b) A current is drawn from the cell until \(1.20 \mathrm{~g}\) of silver has been deposited at the silver electrode. Calculate \(E\) for the cell at this stage of operation.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst equation
To calculate the cell potential under non-standard conditions, the Nernst equation is used: \[ E = E^\circ_{cell} - \frac{RT}{nF} \ln Q \] - **\(E^\circ_{cell}\)** is the standard cell potential. - **\(R\)** is the universal gas constant (8.314 J/mol·K). - **\(T\)** is the temperature in Kelvin. - **\(n\)** is the number of moles of electrons exchanged in the reaction. - **\(F\)** is Faraday's constant (96485 C/mol). - **\(Q\)** is the reaction quotient. Understanding this equation is crucial for determining how variations in concentration affect the voltage of your cell, allowing for the calculation of real-world conditions of a galvanic cell.
Standard electrode potential
Typically, elements with higher standard potentials (more positive) are better at accepting electrons, making them favorable for reduction. Conversely, elements with lower potential values (more negative) tend to lose electrons easily, favoring oxidation.
In a galvanic cell, the net cell potential \(E^\circ_{cell}\) is calculated from the standard potentials of the cathode and anode using: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] This calculation provides the baseline voltage when all conditions are ideal, a vital first step before applying the Nernst equation.
Reaction quotient
In a galvanic cell involving magnesium and silver, the reaction quotient initially is calculated by: \[ Q = \frac{[Mg^{2+}]}{[Ag^+]^2} \] Initially, Q reflects the concentrations of ions before any significant reaction progresses. As the cell operates, the deposition or consumption of ions alters their concentrations, prompting a reevaluation of Q. This adjustment is a key part of predicting changes in the electrochemical potential over time using the Nernst equation.
Silver electrode
Silver ions \((Ag^+)\) in solution get reduced to solid silver \((Ag)\) at the electrode surface. This process results in the deposition of silver metal at the electrodes, as in the given exercise. For every mole of silver reduced, one mole of electrons is transferred, showcasing the simplicity and predictability of silver-based reactions. Understanding how silver electrodes behave is essential when dealing with cell potentials and shifts resulting from changes in concentration.
Magnesium electrode
When magnesium serves as an anode, it releases electrons, destabilizing the metal lattice and forming magnesium ions \((Mg^{2+})\). The process elevates the concentration of these ions in the solution. This metal's high reactivity ensures it delivers significant driving force for electron flow within the galvanic cell.
If you're working on electrochemical experiments, recognizing the role of magnesium is crucial, as it significantly influences the overall cell voltage and facilitates understanding of the cell’s energetics and efficiency.