Question

A student measures the potential of a cell made up with \(1 \mathrm{M}\) CuSO, in one solution and \(1 \mathrm{M} \mathrm{AgNO}_{3}\) in the other. There is a Cu electrode in the CuSO, and an Ag electrode in the AgNO, and the cell is set up as in Figure \(32.1 .\) She finds that the potential, or voltage, of the cell, \(E_{\text {cell }}^{0}\), is \(0.45 \mathrm{V}\), and that the Cu electrode is negative. a. At which electrode is oxidation occurring? b. Write the equation for the oxidation reaction. c. Write the equation for the reduction reaction. d. If the potential of the silver, silver ion electrode, \(E_{A g^{\prime}}^{0}\) angle is taken to be \(0.000 \mathrm{V}\) in oxidation or reduction, what is the value of the potential for the oxidation reaction, \(E_{\mathrm{Cu}, \mathrm{Cu}^{2+} \text { 'arid }}\) ? \(E_{\text {cell }}^{0}=E_{\text {oxid }}^{0}+E_{\text {red }}^{0}\). e. If \(E_{A g^{+} A g \text { red }}^{0}\) equals \(0.80 \mathrm{V}\), as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper, \(E_{\cos a^{2}+a x i d}^{0}\)? f. Write the net ionic equation for the spontaneous reaction that occurs in the cell that the student studied. g. The student adds \(6 \mathrm{M} \mathrm{NH}_{3}\) to be CuSO, solution until the \(\mathrm{Cu}^{2+}\) ion is ecstatically all converted to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) ion. The voltage of the cell, \(E_{\text {cell }}\) goes up to \(0.92 \mathrm{V}\) and the Cu clectrode is still negative. Find the residual concentration of \(\mathrm{Cu}^{2+}\) ion in the cell. (Use Eq. 3.) h. In Part \(g,\left[C u\left(N H_{3}\right)_{4}^{2+}\right]\) is about \(0.05 \mathrm{M},\) and \(\left[\mathrm{NH}_{3}\right]\) is about \(3 \mathrm{M}\). Given those values and the result in Part \(1 \mathrm{g}\) for \(\left[\mathrm{Cu}^{2+}\right],\) calculate \(\mathrm{K}\) for the reaction: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \rightleftarrows \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq})$$

Short answer

The answers to the given questions are as follows: a. Oxidation is occurring at the Cu electrode. b. The oxidation reaction is: \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \). c. The reduction reaction is: \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \). d. The potential for the oxidation reaction of copper is: \(E_{\text{Cu}}^{0} = 0.45 \,V\). e. With the standard reduction potential of silver electrode being 0.80 V, the potential for the oxidation reaction of copper is: \(E_{\text{Cu}^{2}+Cu_{\text{oxid}}}^{0} = -0.35 \,V\). f. The net ionic equation for the spontaneous reaction is: \(\text{Cu} + 2\text{Ag}^{+} \rightarrow \text{Cu}^{2+} + 2\text{Ag} \). g. To find the residual concentration of \(\text{Cu}^{2+}\) ion in the cell, use the Nernst equation as described in the solution. h. To calculate the equilibrium constant, use the formula: \[ K = \frac{[\text{Cu}^{2+}]}{[\text{Cu}\text{(NH}_{3}\text{)}_{4}^{2+}][\text{NH}_{3}]^{4}} \] with the given concentrations and the concentration of \([\text{Cu}^{2+}]\) obtained from part g.

Step-by-step solution

a. Oxidation Location

Oxidation takes place at the anode. Since the Cu electrode is mentioned as being negative, this is where oxidation is occurring.

b. Oxidation Reaction Equation

Oxidation involves losing electrons. As oxidation occurs at the Cu electrode, the equation is: \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \)

c. Reduction Reaction Equation

Reduction takes place at the cathode, which is the Ag electrode in this case. Reduction involves gaining of electrons. So, the equation is: \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \)

d. Potential for Cu Oxidation Reaction

Given that the total cell potential (\(E_{\text{cell}}^0\)), which is 0.45 V, is the sum of the oxidation and reduction potentials, and the potential of the silver electrode (\(E_{\text{Ag}}^{0}\)) is taken to be 0.000 V, the potential of the copper oxidation reaction (\(E_{\text{Cu}}^{0}\)) can be calculated as: \(E_{\text{cell}}^0 = E_{\text{oxid}}^0 + E_{\text{red}}^0\). Therefore, \(E_{\text{Cu}}^{0} = E_{\text{cell}}^0 - E_{\text{Ag}}^{0} = 0.45 \,V - 0.000 \,V = 0.45 \,V\)

e. Potential for Copper Oxidation Reaction with Standard Values

With the standard reduction potential of silver electrode (\(E_{\text{Ag}^{+}Ag_{\text{red}}}^{0}\)) being 0.80 V, and the oxidation-reduction potential relationship \(E_{\text{ext}}^{0}=E_{\text{oxid}}^{0}+E_{\text{red}}^{0}\) still holds, then the potential for the oxidation reaction of copper can be recalculated. Thus, \(E_{\text{Cu}^{2}+Cu_{\text{oxid}}}^{0} = E_{\text{cell}}^{0} - E_{\text{Ag}^{+}Ag_{\text{red}}}^{0} = 0.45 \,V- 0.80 \,V = -0.35 \,V\).

f. Net Ionic Equation

The net equation is obtained by adding the oxidation reaction to the reduction reaction (after balancing for the same number of electrons in both half-reactions). Thus, \(\text{Cu} + 2\text{Ag}^{+} \rightarrow \text{Cu}^{2+} + 2\text{Ag} \)

g. Residual Cu^2+ Concentration

This problem requires the use of the Nernst Equation: \[ E_{cell} = E^{0}_{cell} - \frac{0.0592}{n}logQ \] Where \(Q\) is the reaction quotient. The cell potential \(E_{cell}\) after adding NH3 increases to 0.92V. The number of electrons transferred (\(n\)) is 2 (from the balanced redox reaction). Following the reaction: \[ \text{Cu}^{2+}(aq) + 4\text{NH}_3(aq) \rightarrow \text{Cu}\text{NH}_3^2 +\text{ 2H}_2\text{O} \] The expression for \(Q\) is given as \[\frac{[\text{Cu}^{2+}]^2}{[\text{NH}_3]^4}\] Arrange the Nernst equation to solve for \([\text{Cu}^{2+}]\): \[ [\text{Cu}^{2+}] = \sqrt{\frac{[\text{NH}_3]^4}{10^{n(E^0_{cell} - E_{cell})/0.0592}}} \]

h. Equilibrium Constant Calculation

Given the concentrations of \( \left[C u\left(N H_{3}\right)_{4}^{2+}\right] = 0.05 \,M \), \( \left[NH_{3}\right] = 3 \,M \), and the concentration of \( [\text{Cu}^{2+}] \) obtained from part g, the equilibrium constant (\( K \)) for the reaction can be calculated using the formula: \[ K = \frac{[\text{Cu}^{2+}]}{[\text{Cu}\text{(NH}_{3}\text{)}_{4}^{2+}][\text{NH}_{3}]^{4}} \] This equation is similar to the one used for \( Q \) in part g, but refers to equilibrium concentrations.

Key concepts

Oxidation and Reduction Reactions

Understanding the fundamentals of oxidation and reduction reactions, commonly known as redox reactions, is crucial for comprehending how electrochemical cells function. Oxidation involves the loss of electrons by a species, while reduction involves the gain of electrons.

In an electrochemical cell, these reactions occur at two separate electrodes. The site where oxidation occurs is known as the anode, which becomes negatively charged as electrons are generated. Conversely, the cathode is where reduction happens, acquiring a positive charge due to electron consumption. When a metal like copper (\text{Cu}) gives up electrons to form copper ions (\text{Cu}^{2+}), it is undergoing oxidation:
\[\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-\].
Reducer agents, in this case, such as the silver ions (\text{Ag}^{+}), accept electrons and get reduced in the reaction:
\[\text{Ag}^{+} + e^- \rightarrow \text{Ag}\].The overall process generates a flow of electrons that can be harnessed as electric current, forming the basis of the cell's electrical output.

Nernst Equation

When dealing with non-standard conditions, such as varying concentrations of reactants and products or different temperatures, the Nernst equation provides a way to calculate the potential of an electrochemical cell. The equation is expressed as: \[E_{\text{cell}} = E^{0}_{\text{cell}} - \frac{0.0592}{n}\log Q\],
where \(E_{\text{cell}}\) is the cell potential under non-standard conditions, \(E^{0}_{\text{cell}}\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the half-reactions, and \(Q\) is the reaction quotient, which reflects the ratio of the concentrations of products to reactants at any given moment.
Employing the Nernst equation allows us to understand how the cell potential changes with concentration. As the reaction proceeds and the concentration of reactants decreases while that of products increases, the value of \(Q\) changes, in turn altering the cell potential.

Electrode Potentials

Electrode potentials are intrinsic properties that determine the ability of an electrode to attract or donate electrons during an electrochemical reaction. Every half-reaction has an associated electrode potential, and the difference between the potentials of the two half-cells gives rise to the overall cell potential.

In a cell, the potential at which oxidation occurs at an electrode is matched by an equivalent potential for the reduction at the other electrode, and these together define the voltage of the cell. For instance, the potential for the copper oxidation reaction can be identified when we know the total cell potential and the potential for reduction at the other electrode. This is mathematically portrayed as:\[E_{\text{cell}}^{0}=E_{\text{oxid}}^{0}+E_{\text{red}}^{0}\].
These values are typically measured against a standard hydrogen electrode and compiled in tables. The potential of one half-cell can be derived if the other is known along with the overall potential, which is effectively demonstrated in the textbook problem provided.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a dimensionless number that quantifies the ratio of the concentration of products to reactants at equilibrium in a chemical reaction. It is a vital concept in chemical thermodynamics, offering insight into the direction and extent of chemical reactions.

For the reaction \[\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}(\mathrm{aq}) \rightleftarrows \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq})\],the equilibrium constant is calculated using the formula:\[K = \frac{[\text{Cu}^{2+}]}{[\text{Cu}(\text{NH}_{3})_{4}^{2+}][\text{NH}_{3}]^{4}}\].
This equation is analogous to the formula for the reaction quotient \(Q\) but at equilibrium, the concentrations of reactants and products no longer change. The magnitude of \(K\) can predict whether the reaction favors the formation of products (higher \(K\)) or reactants (lower \(K\)), providing essential information on the reaction's dynamic at equilibrium and allowing us to determine the residual concentration of reactants, as shown in the solution provided for the textbook exercise.