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The cell potential of the following cell at \(25^{\circ} \mathrm{C}\) is \(0.480 \mathrm{~V}\) $$\mathrm{Zn} \mid \mathrm{Zn}^{2+}(1 M) \| \mathrm{H}^{+}(\text {test solution })\left|\mathrm{H}_{2}(1 \mathrm{~atm})\right| \mathrm{Pt}$$ What is the \(\mathrm{pH}\) of the test solution?

Short Answer

Expert verified
The pH of the test solution is 4.745.

Step by step solution

01

Identify the Cell Reaction

The cell is set up such that zinc is oxidized and hydrogen ions in the test solution are reduced. The half-reactions are: 1. Oxidation at the anode: \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^-\)2. Reduction at the cathode: \(2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(g)\)
02

Write the Nernst Equation for the Cell

The Nernst equation at \(25^{\circ} \mathrm{C}\) is given by:\[E_{ ext{cell}} = E^{\circ}_{ ext{cell}} - \frac{0.059}{n} \log Q\]For this reaction, \(n = 2\) because two electrons are transferred. \(Q\) is the reaction quotient, \(Q = \dfrac{[\text{Zn}^{2+}]}{[\text{H}^+]^2} \cdot \dfrac{1}{P_{\mathrm{H}_2}}\). Given that \([\text{Zn}^{2+}] = 1M\) and \(P_{\mathrm{H}_2} = 1 \text{ atm}\), \(Q = \dfrac{1}{[\text{H}^+]^2}\).
03

Calculate Standard Cell Potential

Use the standard electrode potentials: - \(E^{\circ}_{\mathrm{Zn/Zn}^{2+}} = -0.76 \mathrm{~V}\)- \(E^{\circ}_{\mathrm{H}^+/\mathrm{H}_2} = 0.00 \mathrm{~V}\)Thus, the standard cell potential \(E^{\circ}_{\text{cell}} = 0.00 - (-0.76) = 0.76 \mathrm{~V}\).
04

Solve for [H+] Using the Nernst Equation

Substitute the known values into the Nernst equation:\[0.480 = 0.76 - \frac{0.059}{2} \log \left( \dfrac{1}{[\mathrm{H}^+]^2} \right)\]Rearrange and solve for \([\mathrm{H}^+]\):\[0.480 - 0.76 = -0.0295 \log(1/[\mathrm{H}^+]^2)\]\[-0.280 = -0.0295 \log(1/[\mathrm{H}^+]^2)\]\[\log(1/[\mathrm{H}^+]^2) = \frac{0.280}{0.0295}\]\[\log(1/[\mathrm{H}^+]^2) = 9.49\]Solving gives \([\mathrm{H}^+]^2 = 10^{-9.49}\) and \([\mathrm{H}^+] = 10^{-4.745}\).
05

Calculate pH

\(\mathrm{pH} = -\log [\mathrm{H}^+]\)Using the \([\mathrm{H}^+]\) value calculated:\(\mathrm{pH} = -\log(10^{-4.745}) = 4.745\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cell Potential
Cell potential, often denoted as \(E_{\text{cell}}\), is a measure of the potential difference between two electrodes in an electrochemical cell. It gives an indication of how much voltage can be produced by the cell when electrons flow from one electrode to the other.
Two important factors determine cell potential: the nature of the reactants and products, as well as their concentrations. Understanding cell potential helps us predict the direction of electron flow and the feasibility of redox reactions.
The Nernst equation, which comes into play here, adjusts the standard cell potential based on reactant and product concentrations. It helps calculate the actual cell potential for non-standard conditions. The standard condition usually assumes all substances at 1 M concentration and gases at 1 atm pressure, which isn't always the case in practical scenarios.
This makes cell potential calculations crucial in determining the efficiency of cells used in various applications, from batteries to biological systems.
pH Calculation
Calculating pH is essential for understanding the acidity or alkalinity of a solution. The pH scale, ranging from 0 to 14, is a logarithmic scale that represents the concentration of hydrogen ions \([\mathrm{H}^+])\). The formula used is:\[ \mathrm{pH} = - \log [\mathrm{H}^+] \]
In electrochemical cells, such as the one involving Zn and H ions, it is important to calculate pH to understand the electrolyte’s behavior under specific conditions. This leads us back to the Nernst equation, which incorporates pH by altering the hydrogen ion concentration in the reaction quotient \(Q\). By solving the Nernst equation for \([\mathrm{H}^+]\), we can directly calculate the pH of the solution. So, a lower \([\mathrm{H}^+]\) means a higher pH (more basic), while a higher \([\mathrm{H}^+]\) means a lower pH (more acidic).
This relationship underlies many biological and chemical processes, highlighting why accurately calculating and understanding pH is indispensable in scientific fields.
Standard Electrode Potentials
Standard electrode potentials \((E^{\circ})\) are fundamental in understanding redox reactions. They measure the intrinsic tendency of a species to gain or lose electrons, standardized under specific conditions: usually at 25°C, 1 atm pressure, and 1 M concentration.
The standard hydrogen electrode (SHE) is often used as a reference, assigned a potential of 0.00 V. The electrode potentials for other half-reactions are typically measured relative to this.
In the example, the standard electrode potential for the zinc half-reaction, \(E^{\circ}_{\mathrm{Zn/Zn}^{2+}} = -0.76 \mathrm{~V}\) and the hydrogen half-reaction, \(E^{\circ}_{\mathrm{H}^+/\mathrm{H}_2} = 0.00 \mathrm{~V}\), are used. This results in a standard cell potential \(E^{\circ}_{\text{cell}}\) of \(0.76 \mathrm{~V}\), indicating the potential difference under standard conditions.
Understanding standard electrode potentials aids in predicting the feasibility and course of a redox reaction, thereby playing an integral role in the design of batteries and other electrochemical cells. They provide the basis for calculating actual cell potentials using the Nernst equation when conditions deviate from the standard.

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Most popular questions from this chapter

A standard electrochemical cell is made by dipping a silver electrode into a \(1.0 \mathrm{M} \mathrm{Ag}^{+}\) solution and a cadmium electrode into a \(1.0 \mathrm{M} \mathrm{Cd}^{2+}\) solution. a. What is the spontaneous chemical reaction, and what is the maximum potential produced by this cell? b. What would be the effect on the potential of this cell if sodium sulfide were added to the \(\mathrm{Cd}^{2+}\) half-cell and CdS were precipitated? Why? c. What would be the effect on the potential of the cell if the size of the silver electrode were doubled?

Potassium was discovered by the British chemist Humphry Davy when he electrolyzed molten potassium hydroxide. What would be the anode reaction?

Describe what you expect to happen when the following solutions are electrolyzed: a. aqueous \(\mathrm{Na}_{2} \mathrm{SO}_{4} ;\) b. aqueous \(\mathrm{KBr}\). That is, what are the electrode reactions? What is the overall reaction?

A solution of copper(II) sulfate is electrolyzed by passing a current through the solution using inert electrodes. Consequently, there is a decrease in the \(\mathrm{Cu}^{2+}\) concentration and an increase in the hydronium ion concentration. Also, one electrode increases in mass and a gas evolves at the other electrode. Write half-reactions that occur at the anode and at the cathode.

a) Calculate the equilibrium constant for the following reaction at \(25^{\circ} \mathrm{C}\). $$\mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q)$$ The standard cell potential of the corresponding voltaic cell is \(0.030 \mathrm{~V}\). \(\mathrm{b}\) When equal volumes of \(0.75 \mathrm{M}\) solutions of \(\mathrm{Ag}^{+}\) and \(\mathrm{Fe}^{2+}\) are mixed, what is the equilibrium concentration of \(\mathrm{Fe}^{2+} ?\)

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