Question

Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+} ?\) (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

Short answer

In summary, a voltaic cell with the redox reaction of Zn (s) + Sn²⁺ (aq) → Zn²⁺ (aq) + Sn (s) has a standard cell potential of +0.62 V. As the cell operates, the concentration of Zn²⁺ increases while the concentration of Sn²⁺ decreases. When the cell potential drops to zero, the ratio of the concentrations of Zn²⁺ to Sn²⁺ is approximately 1728. The final concentrations when the cell potential drops to zero are approximately 1.499 M for Zn²⁺ and 0.501 M for Sn²⁺.

Step-by-step solution

Calculate the standard cell potential, \(E^\circ\)

To determine \(E^\circ\) for the cell, we need to use the standard reduction potentials for both half-reactions. The half-reactions are: $$\mathrm{Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)} \quad E^\circ_{\mathrm{Sn^{2+}/Sn}} = -0.14\,\text{V}$$ $$\mathrm{Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)} \quad E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76\,\text{V}$$ First, we need to reverse the sign of the reduction potential for the Zn. Since Zn is being oxidized, its reaction should be written as: $$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-} \quad E^\circ_{\mathrm{Zn/Zn^{2+}}} = +0.76\,\text{V}$$ Now that we have both half-reactions correctly written, we can add them together and calculate the cell potential by adding their standard reduction potentials: $$\mathrm{Zn(s) + Sn^{2+}(aq) \rightarrow Zn^{2+}(aq) + Sn(s)} \quad E^\circ_{\text{cell}} = E^\circ_{\mathrm{Zn/Zn^{2+}}} + E^\circ_{\mathrm{Sn^{2+}/Sn}}$$ $$E^\circ_{\text{cell}} = (+0.76\,\text{V}) + (-0.14\,\text{V}) = +0.62\,\text{V}$$

Determine concentration changes during cell operation

As the cell operates, Zn is oxidized and releases \(\mathrm{Zn^{2+}}\) ions into the solution, which will increase the concentration of \(\mathrm{Zn^{2+}}\). At the same time, \(\mathrm{Sn^{2+}}\) ions are reduced and are removed from the solution, decreasing the concentration of \(\mathrm{Sn^{2+}}\).

Find the ratio of concentrations when the cell potential drops

To find the ratio of the concentrations when the cell potential drops to zero, we need to use the Nernst equation: $$E = E^\circ -\frac{RT}{nF}\ln Q$$ Here, \(E\) is the cell potential, \(E^\circ\) is the standard cell potential, R is the gas constant (8.314 J/mol⋅K), T is the temperature in Kelvin, n is the number of electrons transferred, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. When the cell potential drops to zero, the equation becomes: $$0=E^\circ -\frac{RT}{nF}\ln Q$$ For our reaction, we have \(E^\circ=0.62\,\text{V}\), \(n=2\), and \(Q=\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Sn^{2+}}]}\). Rearranging for Q: $$Q=\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Sn^{2+}}]}=e^{(nFE^\circ)/(RT)}$$ Plugging in the values (assuming room temperature of 298 K): $$\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Sn^{2+}}]}=e^{(2\times 96485\,\text{C/mol}\times 0.62\,\text{V})/ (8.314\,\text{J/mol⋅K} \times 298\,\text{K})}$$ $$\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Sn^{2+}}]}=\approx 1728$$

Determine final concentrations

To determine the final concentrations when cell voltage drops to zero, we start with the original concentrations of both cations (\(1.0\,\text{M}\) each). Let \(x\) be the unknown change in concentration for both cations. The change in concentrations can be described as: $$[\mathrm{Zn^{2+}}]_{\text{final}} = 1.0\,\text{M} + x$$ $$[\mathrm{Sn^{2+}}]_{\text{final}} = 1.0\,\text{M} - x$$ Substituting these values into our ratio from step 3: $$\frac{1.0\,\text{M} + x}{1.0\,\text{M} - x} = 1728$$ Solving for \(x\): $$x \approx 0.499\,\text{M}$$ Calculating the final concentrations: $$[\mathrm{Zn^{2+}}]_{\text{final}} \approx 1.0\,\text{M} + 0.499\,\text{M} \approx 1.499\,\text{M}$$ $$[\mathrm{Sn^{2+}}]_{\text{final}} \approx 1.0\,\text{M} - 0.499\,\text{M} \approx 0.501\,\text{M}$$

Key concepts

Standard Cell Potential

Standard cell potential (\(E^\text{o}\)) is crucial in the field of electrochemistry as it represents the voltage difference between two half-cells in a voltaic cell when they are connected by a conductive bridge. A positive value of standard cell potential indicates that the reaction is spontaneous under standard conditions. To calculate this potential, refer to standard reduction potentials which are measured under standard conditions: a temperature of 298 K, a 1 M concentration for each ion participating in the reaction, and a pressure of 1 atm for any gases involved.

To find the overall standard cell potential for a voltaic cell, as in our example, we take the standard reduction potential for the reduction half-reaction, and subtract the standard reduction potential for the oxidation half-reaction (by reversing the sign). The higher the difference in potential, the greater the voltage and the driving force for the electrochemical reaction.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that allows us to calculate the actual cell potential under non-standard conditions. It incorporates the effects of temperature, the number of electrons involved in the electrochemical reaction (n), and the concentrations or partial pressures of the reactants and products (\(Q\)). The Nernst equation is written as:
ewlineewlineewlineewline$$E = E^\text{o} - \frac{RT}{nF}\ln Q$$
ewlineWhere: ewline

• \(E\) = actual cell potential
• \(E^\text{o}\) = standard cell potential
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin
• n = number of moles of electrons transferred in the reaction
• F = Faraday's constant (96485 C/mol)
• Q = reaction quotient, which is the ratio of the products' concentrations to the reactants'

Using the Nernst equation helps us understand how changes in concentration or temperature can affect the cell voltage.

Concentration Changes in Voltaic Cells

When a voltaic cell operates, the concentration of ions in the solutions undergoes changes, due to the redox reactions occurring at the electrodes. As the cell discharges, the anode metal gets oxidized, thus releasing more positive ions into the solution, which increases their concentration. On the contrary, at the cathode, the ions in solution get reduced and deposited onto the electrode, thus decreasing their concentration in the solution.

The analysis of concentration changes is vital because it directly affects the voltage produced by the cell, as indicated by the Nernst equation. In the given exercise, the concentration of ewlineewline\(\text{Zn}^{2+}\) increases while the concentration of ewlineewline\(\text{Sn}^{2+}\) decreases as the cell operates. Understanding these changes can guide us in predicting how long a cell can function before the reactants are depleted.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. Voltaic cells, also known as galvanic cells, are the practical applications of this field, where chemical energy is converted into electrical energy through spontaneous redox reactions. These cells consist of two separate half-cells where oxidation and reduction occur, connected by a salt bridge that maintains charge balance.

The study of electrochemistry encompasses a wide range of topics such as battery technology, corrosion, and electroplating. It aids in the design and improvement of energy storage devices and enables us to predict how different conditions, like temperature and concentration, affect the efficiency and lifespan of these devices. Throughout learning about electrochemistry, one can understand not only the fundamental concepts of how batteries work but also the critical aspects of their maintenance and optimization.