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A student wanted to measure the copper(II) concentration in aqueous solution. For the cathode half-cell she used a silver electrode with a 1.00 -M solution of \(\mathrm{AgNO}_{3}\). For the anode half-cell she used a copper electrode dipped into the aqueous sample. The cell gave \(E_{\text {cell }}=\) \(0.62 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Calculate the copper(II) ion concentration of the solution.

Short Answer

Expert verified
The copper(II) ion concentration is approximately \(3.7 \times 10^{-6}\) M.

Step by step solution

01

Write the Cell Reaction

In this electrochemical cell, the reaction is as follows:\[ \text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s) \quad \text{(Cathode, Reduction)} \]\[ \text{Cu} (s) \rightarrow \text{Cu}^{2+} (aq) + 2e^- \quad \text{(Anode, Oxidation)} \]Combining these half-reactions, the full cell reaction is:\[ 2\text{Ag}^+ (aq) + \text{Cu} (s) \rightarrow 2\text{Ag} (s) + \text{Cu}^{2+} (aq) \]
02

Determine the Standard Electrode Potentials

From standard electrode potential tables, we have:- Reduction potential for silver: \( E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V} \)- Reduction potential for copper: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \)
03

Calculate the Standard Cell Potential

The standard cell potential \( E^\circ_{\text{cell}} \) can be found using the formula:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - 0.34 \text{ V} = 0.46 \text{ V} \]
04

Use Nernst Equation

The Nernst equation for this cell at 25°C (298 K) is:\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln(Q) \]Where \( R = 8.314 \text{ J/mol K} \), \( T = 298 \text{ K} \), \( n = 2 \text{ mol e}^- \), and \( F = 96485 \text{ C/mol e}^- \).\( Q \) is the reaction quotient and in this case is \( \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} \). Since \([\text{Ag}^+]\) is 1.00 M, \( Q = [\text{Cu}^{2+}] \).
05

Substitute Values into Nernst Equation

Given \( E_{\text{cell}} = 0.62 \text{ V} \), substitute all known values into the Nernst equation:\[ 0.62 = 0.46 - \frac{(8.314)(298)}{(2)(96485)} \ln([\text{Cu}^{2+}]) \]\[ 0.62 = 0.46 - 0.0128 \ln([\text{Cu}^{2+}]) \]
06

Solve for \( \ln([\text{Cu}^{2+}]) \)

Rearrange and solve for \( \ln([\text{Cu}^{2+}]) \):\[ 0.62 - 0.46 = -0.0128 \ln([\text{Cu}^{2+}]) \]\[ 0.16 = -0.0128 \ln([\text{Cu}^{2+}]) \]\[ \ln([\text{Cu}^{2+}]) = -12.5 \]
07

Calculate \([\text{Cu}^{2+}]\) Concentration

To find the concentration of \( \text{Cu}^{2+} \), exponentiate both sides:\[ [\text{Cu}^{2+}] = e^{-12.5} \]\[ [\text{Cu}^{2+}] \approx 3.7 \times 10^{-6} \text{ M} \]

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Key Concepts

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

Nernst Equation
The Nernst Equation is a crucial formula in electrochemistry that allows us to calculate the potential of an electrochemical cell under non-standard conditions. It is expressed as:\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln(Q)\]In this equation:
  • \( E_{\text{cell}} \) is the cell potential under the given conditions.
  • \( E^\circ_{\text{cell}} \) signifies the standard cell potential, derived from the standard reduction potentials of the involved half-reactions.
  • \( R \) is the universal gas constant with a value of \( 8.314 \text{ J/mol K} \).
  • \( T \) is the temperature in Kelvin.
  • \( n \) indicates the number of electrons transferred in the cell reaction.
  • \( F \) is the Faraday constant, \( 96485 \text{ C/mol e}^- \).
  • \( Q \) represents the reaction quotient, detailing the concentration or pressures of reactants and products.
By inserting known values into this equation, it's possible to solve for unknowns like the ion concentration, as demonstrated in the exercise to find \([\text{Cu}^{2+}]\).
Through the Nernst Equation, practitioners can understand how various factors influence cell potential outside of standard conditions.
Standard Electrode Potential
Standard Electrode Potential, denoted by \( E^\circ \), is the voltage achieved when a half-cell is connected to a standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C).
Each half-reaction's standard electrode potential is determined and used to predict the cell’s overall standard potential.For instance, in the given exercise, the standard electrode potential for silver, \( E^\circ_{\text{Ag}^+/\text{Ag}} \), is \( +0.80 \text{ V} \).
Meanwhile, copper’s standard electrode potential, \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \), is \( +0.34 \text{ V} \). These values are derived from experimental data and compiled in reference tables.
To find the standard cell potential \( E^\circ_{\text{cell}} \), the formula utilized is:\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]This simplifies understanding the tendency of a redox reaction to proceed and its electrical potential difference under ideal conditions.
Grasping standard electrode potentials is vital, as they serve as building blocks to analyze more complex electrochemical systems.
Reaction Quotient
The Reaction Quotient, represented as \( Q \), plays a central role in thermodynamics and electrochemistry. It is used to gauge the direction a reaction must go to reach equilibrium.
For a reaction:\[aA + bB \rightleftharpoons cC + dD\]The reaction quotient \( Q \) is calculated as:\[Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]Here, \([A]\), \([B]\), \([C]\), and \([D]\) are the molar concentrations of the reactants and products. In scenarios where the reaction involves gases, partial pressures are used.
In the context of the provided exercise, the reaction quotient was simplified to \( Q = [\text{Cu}^{2+}] \) because the concentration of silver (\([\text{Ag}^+]\)) was given as 1 M.
As the Nernst Equation is dependent on \( Q \), understanding its value helps determine shifts from expected standard conditions, revealing the concentration or pressure changes needed for equilibrium.
Reaction Quotient analysis is quintessential in predicting and interpreting the behavior of chemical reactions in dynamic systems.

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Most popular questions from this chapter

Balance these redox reactions, and identify the oxidizing agent and the reducing agent. (a) \(\mathrm{FeO}(\mathrm{s})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})\) (b) \(\mathrm{P}_{4}(\mathrm{~s})+\mathrm{Br}_{2}(\ell) \longrightarrow \mathrm{PBr}_{5}(\ell)\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Co}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Co}^{3+}(\mathrm{aq})\) in acidic solution (d) \(\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{Cr}^{3+}(\mathrm{aq})\) in acidic solution (e) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s})\) in basic solution (f) \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (g) \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

You wish to electroplate a copper surface having an area of \(1200 \mathrm{~mm}^{2}\) with a \(1.0-\mu \mathrm{m}\) -thick coating of silver from a solution of \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) ions. If you use a current of \(150.0 \mathrm{~mA},\) calculate how much electrolysis time you should use. The density of metallic silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3} .\)

Assume that the anode reaction for the lithium battery is $$ \mathrm{LiC}_{6}(\mathrm{~s}) \longrightarrow \mathrm{Li}^{+}(\text {electrolyte })+\mathrm{C}_{6}(\mathrm{~s})+\mathrm{e}^{-} $$ and the anode reaction for the lead-acid storage battery is $$ \mathrm{Pb}(\mathrm{s})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \longrightarrow \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{e}^{-}+\mathrm{H}^{+}(\mathrm{aq}) $$ Compare the masses of metals consumed when each of these batteries supplies a current of \(1.0 \mathrm{~A}\) for \(10 . \mathrm{min}\).

An electrolytic cell is set up with \(\mathrm{Cd}(\mathrm{s})\) in \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(\mathrm{Zn}(\mathrm{s})\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) .\) Initially both electrodes weigh \(5.00 \mathrm{~g}\). After running the cell for several hours the electrode in the left compartment weighs \(4.75 \mathrm{~g}\). (a) Which electrode is in the left compartment? (b) Does the mass of the electrode in the right compartment increase, decrease, or stay the same? If the mass changes, what is the new mass? (c) Does the volume of the electrode in the right compartment increase, decrease, or stay the same? If the volume changes, what is the new volume? (The density of \(\mathrm{Cd}\) is \(\left.8.65 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

To measure the \(\mathrm{Ag}^{+}\) concentration, \(25.00 \mathrm{~mL}\) of a silvercontaining solution is titrated with \(0.015 \mathrm{M} \mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) by using a silver electrode immersed in the test solution and the electrical potential measured against a standard hydrogen electrode. It required \(16.7 \mathrm{~mL}\) of the KI solution to reach the equivalence point, where the potential was \(0.325 \mathrm{~V}\). (a) Calculate the molarity of \(\mathrm{Ag}^{+}\) in the solution. (b) Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{AgI}\).

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