Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q)+2 \mathrm{Au}(s) \longrightarrow 2 \mathrm{Au}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O} $$ (a) Calculate \(E\). (b) Write the Nernst equation for the cell. (c) Calculate \(E\) when \(\left[\mathrm{Au}^{3+}\right]=0.250 \mathrm{M},\left[\mathrm{H}^{+}\right]=1.25 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=\) \(1.50 M\)

Short Answer

Expert verified
Answer: The cell potential, E, under the given concentrations is approximately -0.048 V.

Step by step solution

01

a) Calculate E

Step 1: Write half-reactions $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) + 2e^- \longrightarrow 2\mathrm{H}_{2} \mathrm{O} $$ $$ 2 \mathrm{Au}(s) + 6e^- \longrightarrow 2 \mathrm{Au}^{3+}(a q) $$ Step 2: Find standard reduction potentials of half-reactions We look up the standard reduction potentials (E°) for each half-reaction from a table. We find that: $$ E°_\text{H2O2} = 1.776\,\mathrm{V} $$ $$ E°_\text{Au} = 1.498\,\mathrm{V} $$ Step 3: Calculate E The standard cell potential (E) can be calculated using the equation: $$ E = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} $$ Using H2O2 as the anode and Au as the cathode, the equation becomes: $$ E = E^{\circ}_{\text{Au}} - E^{\circ}_{\text{H2O2}} $$ Substituting the values, we can find E: $$ E = 1.498 - 1.776 $$ $$ E = -0.278\,\mathrm{V} $$
02

b) Write the Nernst equation for the cell

The Nernst equation relating cell potential (E) with concentration is given by: $$ E = E^\circ - \frac{RT}{nF}\ln Q $$ Where R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient. The given reaction has: $$ n = 6 $$ $$ Q = \frac{[\mathrm{Au}^{3+}]^2[H_{2}O]^6}{[H_{2}O_{2}]^3[H^+]^6} $$ Substituting the values for R, T, n, and F (with T=298K and 25°C and F=96485 C/mol), we get: $$ E = E^\circ - \frac{8.314\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1} \times 298\,\text{K}}{6 \times 96485\,\text{C}\,\text{mol}^{-1}}\ln Q $$ Simplifying, $$ E = E^\circ - 0.00831\,\mathrm{V}\ln Q $$
03

c) Calculate E when given concentrations

Using the concentrations given in the problem and the reaction quotient Q, $$ Q = \frac{(0.250)^2\cdot(1)^6}{(1.50)^3\cdot(1.25)^6} $$ Now, we can use the Nernst equation to calculate the cell potential E: $$ E = -0.278 - 0.00831\,\mathrm{V}\ln Q $$ Substituting the value of Q, $$ E = -0.278 - 0.00831\,\mathrm{V}\ln \left(\frac{(0.250)^2\cdot(1)^6}{(1.50)^3\cdot(1.25)^6}\right) $$ After calculating the answer, $$ E \approx -0.048\,\mathrm{V} $$

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Key Concepts

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

Standard Reduction Potential
The standard reduction potential, commonly represented as E°, is a measure of the tendency of a chemical species to acquire electrons and thus be reduced. Each half-reaction in an electrochemical cell has an associated standard reduction potential, which is measured under standard conditions: a temperature of 298 K, a 1 M concentration for each aqueous species, and a pressure of 1 atm for any gases involved.

In the context of our voltaic cell example, the standard reduction potentials for the oxidation of hydrogen peroxide and the reduction of gold were looked up from a reference table. These values are essentially guides that tell us how readily these reactions occur compared to the standard hydrogen electrode which is assigned an E° of 0 V. A higher standard reduction potential indicates a greater tendency to gain electrons and be reduced, whereas a lower value signifies a tendency to lose electrons. Understanding E° is crucial for determining the overall cell potential and predicting the direction of the reaction.
Nernst Equation
The Nernst equation provides a relationship between the cell potential at any given condition and the standard cell potential, E°. The equation takes into account the effect of temperature and the concentrations of the reacting species. It can be represented as:
\[ E = E^\circ - \frac{RT}{nF}\ln Q \]
where:
  • \(E\) is the cell potential under non-standard conditions,
  • \(E^\circ\) is the standard cell potential,
  • \(R\) is the universal gas constant (8.314 J/mol·K),
  • \(T\) is the temperature in kelvins,
  • \(n\) is the number of moles of electrons exchanged,
  • \(F\) is Faraday's constant (96,485 C/mol),
  • \(Q\) is the reaction quotient, which is the ratio of the products' activities to the reactants' activities.

By using the Nernst equation, you can calculate the cell potential for any given set of conditions, taking into account the concentrations of the reactants and products. This is valuable for predicting the behavior of the cell under a variety of real-world scenarios.
Cell Potential Calculation
Calculating the cell potential, often designated as E, is a key task in electrochemistry. For a standard cell potential, the calculation is straightforward and involves the difference between the cathode (reduction) and anode (oxidation) standard reduction potentials:
\[E = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]
However, in practical scenarios, the conditions may not be standard. Concentrations can differ from the 1 M standard. In such cases, we use the Nernst equation to find the actual cell potential. By plugging in the values for temperature, the number of electrons transferred (n), and the reaction quotient (Q), you can adjust the standard cell potential to find the potential at these specific non-standard conditions.

The example provided in the exercise demonstrates this calculation, showing how to use the Nernst equation to adjust the standard cell potential based on the given concentrations of the species involved in the reaction.
Electrochemistry
Electrochemistry is a branch of chemistry that deals with the study of the interrelation of electrical and chemical phenomena. It is primarily concerned with how chemical reactions can generate electricity, as in voltaic or galvanic cells, and how electrical energy can be used to cause chemical transformations, which happens in electrolytic cells.

In a voltaic cell, a spontaneous chemical reaction generates an electric current. This process involves the transfer of electrons from one chemical species to another through an external circuit. The movement of ions within the cell completes the circuit. The standard reduction potentials, which we have discussed, are fundamental to understanding which reactions can produce electrical current, and are used in conjunction with the Nernst equation to calculate the exact cell potential under any given conditions.

The fundamental principles of electrochemistry like oxidation-reduction (redox) reactions, cell potentials, and the use of the Nernst equation are not just academic exercises. They are vital in the development of batteries, fuel cells, corrosion prevention, and many other technological applications that play a crucial role in our daily lives.

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Most popular questions from this chapter

Given the following standard reduction potentials $$ \begin{gathered} \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) \quad E^{\circ}=0.799 \mathrm{~V} \\ \mathrm{Ag}(\mathrm{CN})_{2}^{-}+e^{-} \longrightarrow \mathrm{Ag}(s)+2 \mathrm{CN}^{-}(a q) \quad E^{\circ}=-0.31 \mathrm{~V} \end{gathered} $$ find \(K_{f}\) for \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)\) at \(25^{\circ} \mathrm{C}\).

Consider a salt bridge voltaic cell represented by the following reaction: $$ \mathrm{Fe}(s)+2 \mathrm{Tl}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Tl}(s) $$ Choose the best answer from the choices in each part below: (a) What is the path of electron flow? Through the salt bridge, or through the external circuit? (b) To which half-cell do the negative ions in the salt bridge move? The anode, or the cathode? (c) Which metal is the electrode in the anode?

An alloy made up of tin and copper is prepared by simultaneously electroplating the two metals from a solution containing \(\mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). If \(20.0 \%\) of the total current is used to plate tin, while \(80.0 \%\) is used to plate copper, what is the percent composition of the alloy?

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?

What is \(E^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction? $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s) $$ \(K_{s p}\) for \(\mathrm{BaSO}_{4}\) is \(1.1 \times 10^{-10}\)

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