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Calculate the voltage (E) of a cell with \(\mathrm{E}^{\circ}=1.1\) volts, If the copper half-cell is at standard conditions but the zinc ion concentration is only \(.001\) molar. Temperature is \(25^{\circ} \mathrm{c}\). The overall reaction is \(\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Cu}+\mathrm{Zn}^{+2}\)

Short Answer

Expert verified
The voltage (E) of the cell can be calculated using the Nernst equation, given the standard voltage (\(E^0\)) of 1.1 volts, temperature of \(25^\circ \mathrm{C}\) and knowing that the copper half-cell is at standard conditions while the zinc ion concentration is 0.001 M. After converting temperature to Kelvin, determining the number of moles of electrons transferred in the overall reaction, and finding the reaction quotient (Q), we can plug all the values into the Nernst equation and calculate the cell voltage (E). The voltage (E) of the cell is approximately 1.19 volts.

Step by step solution

01

Recall the Nernst equation

The Nernst equation helps us relate the cell potential (E) to the standard cell potential \((E^0)\), the temperature (T), the number of moles of electrons transferred (n), the reaction quotient (Q), and the gas constant (R) and Faraday's constant (F). The Nernst equation in a simplified form is: \[E = E^0 - \frac{RT}{nF} \ln Q\] In this problem, we are given \(E^0\), and we need to find E. To do this, we need to determine the temperature T in Kelvin, the number of moles of electrons (n) transferred in the reaction, and the reaction quotient (Q).
02

Convert temperature to Kelvin

Given the temperature is 25°C, convert it to Kelvin (K) using the relation: \[T(K) = T(°C) + 273.15\] Plugging in the given values: \[T(K) = 25 + 273.15 = 298.15\ K\]
03

Determine the number of moles of electrons transferred

In the balanced redox equation, Zn + Cu^2+ → Cu + Zn^2+, we can see that Zn loses 2 electrons (oxidation) and Cu^2+ gains 2 electrons (reduction). Therefore, the number of moles of electrons transferred (n) in the reaction is 2. \[n = 2\]
04

Calculate the reaction quotient (Q)

The reaction quotient (Q) for the given redox reaction is expressed as: \[Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}\] The copper half-cell is at standard conditions, which means the concentration of Cu^2+ is 1 M. The zinc ion concentration is given as 0.001 M. Plug these values into the expression for Q: \[Q = \frac{0.001}{1} = 0.001\]
05

Apply the Nernst equation

Now we have all the values needed to plug into the Nernst equation: \[E = E^0 - \frac{RT}{nF} \ln Q\] Use the gas constant R = 8.314 J/(mol·K), Faraday's constant F = 96485 C/mol, and the values calculated in previous steps: \[E = 1.1 - \frac{8.314 \times 298.15}{2 \times 96485} \ln 0.001\] Calculate the value inside the ln function: \[\ln 0.001\approx-6.91\] Now plug this value into the equation: \[E = 1.1 -\frac{8.314 \times 298.15}{2 \times 96485} \times (-6.91)\] \[E = 1.1 + 0.0896\] Therefore, the voltage (E) of the cell is approximately: \[E \approx 1.19\ \text{volts}\]

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Key Concepts

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

Electrochemistry
Electrochemistry is a branch of chemistry that deals with the interrelation of electrical currents and chemical reactions. It's a study of how chemicals produce electricity and how electricity leads to chemical changes. This area of chemistry is important for a wide range of applications, including batteries, fuel cells, and electrolysis processes.

Within an electrochemical cell, chemical energy is converted into electrical energy through redox reactions, involving the transfer of electrons. The two key components making up an electrochemical cell are the anode, where oxidation (loss of electrons) occurs, and the cathode, where reduction (gain of electrons) takes place. The overall chemical reaction in a cell is a redox reaction, resulting from the combined reduction and oxidation processes happening at the electrodes.
Cell Potential
Cell potential, often denoted as E, is a measure of the driving force behind the electrical current flow in an electrochemical cell. It is a quantitative expression of the cell's ability to do electric work and is measured in volts (V). The cell potential results from the difference in the potential energy of electrons at the anode and cathode.

The cell potential can be further understood by considering it as the voltage that develops between the two electrodes when they are connected through an external circuit. If the cell potential is positive, the reaction is spontaneous. Conversely, if it's negative, the reaction will not proceed without external electrical energy (non-spontaneous). Understanding cell potential is vital in determining the efficiency and feasibility of electrochemical reactions.
Reaction Quotient
The reaction quotient, Q, represents the relative amounts of products and reactants involved in a reaction at a given moment. For a redox reaction, it is expressed as the ratio of the concentrations of the products raised to the power of their stoichiometric coefficients to those of the reactants raised to the power of their stoichiometric coefficients.

The reaction quotient is a crucial concept in predicting the direction in which a reaction will proceed to reach equilibrium. If Q is less than the equilibrium constant (K), the forward reaction is favored, and if Q is greater than K, the system will shift to favor the reverse reaction. In the context of electrochemistry, the reaction quotient is an essential factor in the Nernst equation, influencing the cell potential.
Standard Cell Potential
The standard cell potential, denoted as \(E^0\), is the voltage or electrical potential difference of a cell under standard conditions, which includes solute concentrations of 1 mol/L, a pressure of 1 bar for any gases involved, and a specified temperature, usually 25°C. The standard cell potential is calculated by subtracting the standard potential of the cathode from the anode.

Since it provides a benchmark to compare the innate voltage potential of cells, \(E^0\) is a key reference for determining reaction spontaneity and direction as well. It is also used as a starting value in the Nernst equation for calculating the cell potential under non-standard conditions.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, are chemical processes in which electrons are transferred between two substances. The substance that loses electrons is said to be oxidized while the one gaining electrons is reduced. There are two half-reactions in a redox process: one for oxidation and one for reduction.

Redox reactions are the foundation of electrochemical cells – in a cell, the oxidation occurs at the anode, and the reduction occurs at the cathode. These coupled reactions are pivotal in converting chemical energy into electrical energy, powering devices ranging from handheld batteries to industrial-scale power plants. Understanding the balancing of redox reactions is critical in designing and analyzing electrochemical cells.

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Most popular questions from this chapter

Show how Le Chatelier's principle for oxidation-reduction reactions corresponds to the Nernst equation.

For the following oxidation-reduction reaction, (a) write out the two half- reactions and balance the equation, (b) calculate \(\Delta E^{\circ}\), and (c) determine whether the reaction will proceed spontaneously as written; \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\)

(a) You desire to plate 5 g of Ag from a solution of \(\mathrm{Ag}^{+}\). Determine the number of amperes required over a period of 1000 seconds. (b) With a current of \(0.5\) ampere, how much time is required to plate out \(10 \mathrm{~g}\) of gold from an \(\mathrm{Au}^{3+}\) solution?

A voltaic cell is made up of a silver electrode in a \(1 \mathrm{M}\) silver nitrate solution and an aluminum electrode in a \(1 \mathrm{M}\) aluminum nitrate solution. The half reactions are 1) \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\) (s) \(\mathrm{E}^{\circ}=.80\) volt, and (2) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s})\) \(E^{\circ}=-1.66\) volt, calculate the standard free energy change, \(\Delta \mathrm{G}^{\circ}\), in \(\mathrm{Kcal} /\) mole. \(1 \mathrm{cal}=4.184\) joule \(1 \mathrm{~F}=96,500\) coul, \(1 \mathrm{Kcal}=1000 \mathrm{cal}\)

Construct a galvanic cell based on the reaction \(3 \mathrm{Fe}(\mathrm{s})+2 \mathrm{Au}^{+3} \rightarrow 3 \mathrm{Fe}^{+2}+2 \mathrm{Au}(\mathrm{s})\) with \(\Delta \mathrm{H}\) negative.

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