Extra Credit 21

Q17.2.10

The mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).

S17.2.10

A typical cell diagram is given like this \( \text{Pt}(s)| \text{H}_2 (g) | \text{OH}^-(aq) || \text{Cl}^-(aq)|\text{Cl}_2(g) | \text{Pt}(s)\) where each line \( | \) represents a phase change in the cell. The double line \( ||\) indicates the presence of a salt bridge to complete the circuit and maintain the flow of electrons. In this particular case, the electrodes are both solid platinum and are not participating in the reaction; the electrodes are inert and are simply there to allow for flow of electrons in the cell compartments.

\[ {H_2}_{(g)} + Cl^-_{(aq)} \to OH^-_{(aq)} + {Cl_2}_{(g)}\]

In other cases, it is possible that the electrodes can take part in the reaction and often directly interact with the components of the half reactions. For example, the cell diagram given by \( \text{Ni}(s)| \text{Ni}^{2+}(aq) || \text{Ag}^+(aq)|\text{Ag}(s) \) indicates that the nickel electrode and silver electrode both partake in the reactions as described below.

\[ {Ni}_{(s)} + Ag^+_{(aq)} \to Ni^+_{(aq)} + {Ag}_{(s)}\]

The interesting thing about the case above is that the half reactions involve turning one aqueous ion into its solid form. This process is known as electroplating, where a metal is deposited onto another substance through its half reaction. In terms of our problem, it is clear that one electrode is gaining mass where an ion is being plated onto the electrode, while another electrode is losing mass since the atoms of the electrode are turning into aqueous ions through the cell's half reaction.

\[ \textbf{Oxidation/Anode: } {Ni}_{(s)} \to Ni^+_{(aq)} + e^- \]

\[ \textbf{Reduction/Cathode: } Ag^+_{(aq)} + e^- \to {Ag}_{(s)} \]

From the above reactions we can determine that the cathode gains mass through the reduction half reaction, while the anode loses masses through the oxidation half reaction. The inert electrode does not lose nor gain mass during the reactions and is not directly active. (The inert electrode does not change mass because it acts as a medium to deliver the electron to the Cl^- species in the solution. It is also worth mentioning that platinum is a good inert electrode because it is active and stable.) -Yvanna Todorova

\( \textbf{Active: }A, C \qquad \textbf{Inactive: }B\)

\( \textbf{Cathode: }A \qquad \textbf{Anode: }C\)

Q19.3.11

Would you expect the Mg₃[Cr(CN)₆]₂ to be diamagnetic or paramagnetic? Explain your reasoning.

S19.3.11

In order to solve this problem we must first deconstruct the components of the coordination complex. Coordination complexes typically consist of a transition metal element that is called the central atom. The compounds directly bonded to the central atom through covalent bonds are called ligands. This particular compound tells us that there are 6 cyanide ligands bound to the central transition element, chromium. There are also 3 magnesium counter ions bound to the anion coordination sphere. A counter ion is an ion that is bound to the coordination complex through an ionic bond rather than a covalent bond, and is considered to be outside of the coordination sphere.

In this case, each cyanide ligand has a charge of -1, resulting in a total charge of -6. Each of the three magnesium ions has a charge of +2, resulting in a total charge of +6. We must now determine the charge on the central chromium atom with the knowledge that the overall charge of this complex is 0 (Wrong: You would want the charge of the molecule to equal -3. The three Magnesium create a total charge of +6 so the complex needs to equal -6 but since there are two, each should have a charge of -3. The six cyanide ligands create a -6 charge so Cr must have a +3 charge to create an overall charge of -3) -Yvanna Todorova.

If we let \(\textbf{x = charge of the Chromium ion} \dots \)

\[ \textbf{2x} \text{ + 3(+2) + 12(-1) = 0} \]

\[ \text{x = +3} \implies \text{oxidation state of Cr is +3} \]

Now that we know the oxidation state of Chromium, we can determine the number of d-electrons in the Cr³⁺ central ion. The typical electron configuration for chromium is

\[\text{[Ar}]4\text{s}^23\text{d}^4 \]

The Cr³⁺ ion has 3 fewer electrons that are removed from the highest energy levels first, starting with the 4s subshell.

\[\text{[Ar}]3\text{d}^3 \]

The Cr³⁺ ion loses three electrons starting from the 4s subshell, leaving it with \(\textbf{3 d-electrons}\). Next, we must determine whether the cyanide ligand is a strong field or weak field ligand in order to predict the spin of the compound. We will use the spectrochemical series to determine that CN^- is a strong field ligand, which in turn produces a low spin complex(It is low spin because the Delta o in strong field ligands is larger than the crystal field splitting energy) -Yvanna Todorova. The six monodentate cyanide ligands indicate that the complex has a coordination number of 6 and is octahedral. The crystal field diagram is constructed below.

The diagram indicates that there are 2 unpaired electrons resulting in a \(\textbf{paramagnetic complex}. \)

Q12.4.11

The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?

S12.4.11

\[ \textbf{Initial Conditions: } t_{1/2} = 8.50 \text{ min} \quad A_0 = 0.150 \text{ mol/L} \]

\[ \textbf{First Order Reaction: }t_{1/2} = \frac{ln2}{k} = \frac{0.693}{k} \qquad \textbf{Second Order Reaction: }t_{1/2} = \frac{1}{k [A]_0} \]

In order to solve this problem we will use the exponential decay function \( A = A_0e^{-kt} \) and first determine our k constant for both the first order and second order reactions. In case (a) the reaction is first order with respect to A and we will use \( t_{1/2} = \frac{ln2}{k} \) to determine that \[ \textbf{k} = \frac{ln2}{t_{1/2}} = \frac{ln2}{8.5min * 60\frac{sec}{min}} = \textbf{1.359 x 10}^{\textbf{-3}}s^{-1} \]

Now that we have determined the k constant for the first order reaction, we can use the exponential decay equation to calculate the amount of time t it will take for the concentration of A to drop to 0.0300 mol/L.

(I would have used the ln[A]_t=-kt=ln[A]_o but they get the same answer so both ways work) -Yvanna Todorova

\[ A = A_0e^{-kt} \]

\[ \text{0.0300 mol/L }= \text{0.150 mol/L} \bullet \text{e}^{\text{-1.359 x 10}^{-3}\bullet \text{ t }} \]

\[ ln\frac{0.0300}{0.150} = {\text{-1.359 x 10}^{-3}\bullet \text{ t }} \bullet ln \text{ e} \]

\[ \textbf{ t = 1184.1 sec } \quad OR \quad \textbf{ 19.73 min} \]

In part (b) the reaction is second order with respect to A so we will use the same process, except with the half life equation for second order reaction \(t_{1/2} = \frac{1}{k [A]_0}\), to determine that \[ \textbf{k} = \frac{1}{t_{1/2}[A]_0} = \frac{1}{8.5min * 60\frac{sec}{min} * 0.150mol/L } = \textbf{1.307 x 10}^{\textbf{-2}}s^{-1} \]

Now that we have determined the k constant for the second order reaction, we can use the exponential decay equation to calculate the amount of time t it will take for the concentration of A to drop to 0.0300 mol/L.

(Once again, another way to calculate the answer is 1/[A]_t=kt+1/[A]_o but both methods yield the correct answer)-Yvanna Todorova

\[ \frac{1}{A} = \frac{1}{A_0} + kt \]

\[ \frac{1}{\text{0.0300 mol/L }}= \frac{1}{\text{0.150 mol/L}} + \text{1.307 x 10}^{-2}\bullet \text{ t } \]

\[ \frac{80}{3} = \text{1.307 x 10}^{-2}\bullet \text{ t } \]

\[ \textbf{ t = 2571.5 sec } \quad OR \quad \textbf{ 42.86 min} \]

Q21.2.6

Calculate the density of the \( ^{24}_{12}Mg\) nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10^–13 cm and is spherical in shape.

S21.2.6

\[ \textbf{Initial Conditions: } V_{\text{sphere}} = \frac{4}{3}\pi \text{r}^3 \qquad radius = \frac{diameter}{2} = \text{5 x 10}^{-14} cm\]

We are asked to determine the density of the Mg-24 nucleus, given the diameter and approximate shape. First, we need to realize that density is calculated as \( \frac{mass}{volume}\) where the mass of the Mg nucleus should be in grams and the volume in \( cm^3\) or mL (It is important to understand this relationship because it is not stated in the problem. This is a good example of the importance of paying attention to units)-Yvanna Todorova. The mass of the Mg nucleus can be calculated by determining the number of protons and neutrons present. The atomic number of Mg is 12, indicating that there are 12 protons in this nucleus and \(\text{24 - 12 = 12 neutrons} \). The mass of the nucleus is calculated to be:

\[ \text{Total Mass = } \text{ # of protons} \bullet \frac{\text{1.6727 x 10}^{-24} }{1 proton} + \text{# of neutrons} \bullet \frac{\text{1.6750 x 10}^{-24} }{1 neutron}\]

\[ \textbf{Total Mass = 4.017 x 10}^{-23} g \]

The volume of the nucleus can be calculated using the formula for the volume of a sphere. The radius is equal to \( \frac{1}{2}diameter\).

\[ \text{Volume of Nucleus = } \frac{4}{3} \pi r^3\]

\[ \text{Volume of Nucleus = } \frac{4}{3} \pi (\text{5 x 10}^{-14}cm)^3\]

\[ \text{Volume of Nucleus = } \text{5.236 x 10}^{-40} cm^3\]

The density can be determined after calculating the total mass and total volume:

\[ \text{Density = } \frac{mass}{volume} \]

\[ \text{Density = } \frac{\text{4.017 x 10}^{-23} g}{\text{5.236 x 10}^{-40} cm^3} \]

\[ \textbf{Density = 7.672 x 10}^{16} \frac{g}{mL} \]

Q21.6.1

How can a radioactive nuclide be used to show that the equilibrium: \( AgCl(s)⇌Ag^+(aq)+Cl^−(aq)\) is a dynamic equilibrium?

S21.6.1

This equation shows that solid silver chloride dissociates into aqueous silver ion and aqueous chloride ion at equilibrium. We can see that the double arrows \( \iff \) indicate that the reaction is continuously going forwards and backwards. Although the concentrations of the products and reactants do not necessarily have to be equal, the rates of the forward and reverse reactions are the same at equilibrium. Thus, introduction of either radioactive silver ion, Ag⁺, or radioactive chloride ion, Cl^–, into the solution containing the above reaction will produce a radioactive precipitate, specifically \( AgCl_{(s)} \) as the reverse reaction takes place.(I agree with this answer, in addition, these relations come from le chatelier's principle about equilibrium)-Yvanna Todorova

If we let \( {\color{red}\text{ Ag}^+} = \text{ radioactive silver} \dots \)

\[ {\color{red}\text{ Ag}^+_{(aq)}} + Cl^-_{(aq)} \iff {\color{red}\text{ Ag}}Cl_{(s)} \]

Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

1. Cr(s) + Ni²⁺(aq) → Cr²⁺(aq) + Ni(s)
2. Cl₂(g) + Sn²⁺(aq) → 2Cl⁻(aq) + Sn⁴⁺(aq)
3. H₃AsO₄(aq) + 8H⁺(aq) + 4Zn(s) → AsH₃(g) + 4H₂O(l) + 4Zn²⁺(aq)
4. 2NO₂(g) + 2OH⁻(aq) → NO₂⁻(aq) + NO₃⁻(aq) + H₂O(l)

S20.4.7

In order to determine the oxidant and reductant for each of the above reactions, we must break down the overall reactions into their corresponding half reactions. Then, we will determine the oxidation numbers of each element and track the number of electrons transferred to conclude whether a particular element is an oxidant or reductant. To clarify, an oxidant or oxidizing agent is the ion or compound that allows for the oxidation of another element, and is itself reduced. In similar terms, a reductant or reducing agent is the ion or compound responsible for the reduction of another element in the reaction, and is itself oxidized.

\(\textbf{1).}\) For reaction 1 we see that the overall reaction \( Cr_{(s)} + Ni^{ 2+}_{(aq)} \to Cr^{ 2+}_{(aq)} + Ni_{(s)} \) can be broken down into the following half reactions:

\[ Cr_{(s)} \to Cr^{ 2+}_{(aq)} \qquad Ni^{ 2+}_{(aq)} \to Ni_{(s)}\]

We see that the oxidation number for solid Cr is 0 on the reactants side, but +2 on the products side for \(Cr^{ 2+} \). Similarly, the oxidation number for \(Ni^{ 2+} \) is +2 on the reactants side and 0 on the products side for solid Ni. From the half reactions we can determine that Cr is oxidized as it loses electrons and gains a positive charge (i.e. 0 to +2) while Ni is reduced as it gains electrons and acquires a negative charge (i.e. +2 to 0). Now, we can conclude:

\( \textbf{Oxidant: } Ni^{ 2+}_{(aq)} \qquad \textbf{Reductant: } Cr_{(s)} \)

\(\textbf{2).}\) For reaction 2 we see that the overall reaction \( Cl_{2(g)} + Sn^{ 2+}_{(aq)} \to 2Cl^{ -}_{(aq)} + Sn^{4+}_{(aq)} \) can be broken down into the following half reactions:

\[ Cl_{2(g)} \to 2Cl^{ -}_{(aq)} \qquad Sn^{ 2+}_{(aq)} \to Sn^{4+}_{(aq)}\]

We see that the oxidation number for gaseous Cl is 0 on the reactants side, but -1 on the products side for \(2Cl^{ -}_{(aq)} \). Similarly, the oxidation number for \(Sn^{ 2+}_{(aq)} \) is +2 on the reactants side and +4 on the products side for aqueous Sn. From the half reactions we can determine that \(Sn^{ 2+} \) is oxidized as it loses electrons and gains a positive charge (i.e. +2 to +4) while Cl is reduced as it gains electrons and acquires a negative charge (i.e. 0 to -1). Now, we can conclude:

\( \textbf{Oxidant: } Cl_{2(g)} \qquad \textbf{Reductant: } Sn^{ 2+}_{(aq)} \)

\(\textbf{3).}\) For reaction 3 we see that the overall reaction \( H_3AsO_{4(aq)} + 8H^{ +}_{(aq)} + 4Zn_{(s)} \to AsH_{3(g)} + 4H_2O_{(l)} + 4Zn^{ 2+}_{(aq)} \) can be broken down into the following half reactions:

\[ H_3AsO_{4(aq)} + 8H^{ +}_{(aq)} \to AsH_{3(g)} + 4H_2O_{(l)} \qquad 4Zn_{(s)} \to 4Zn^{ 2+}_{(aq)}\]

Since we know that the oxidation number for an oxygen in a compound is typically -2 and that the oxidation number for hydrogen is +1. We see that the oxidation number for As is +5 on the reactants side, but -3 on the products side for \(AsH_{3(g)} \).

\[ \textbf{Reactants} \qquad \qquad \textbf{ Products}\]

\[ \text{3(+1) + (oxidation state As) + 4(-2) = 0} \qquad \text{3(+1) + (oxidation state As) = 0}\]

\[ \text{oxidation state As = +5} \qquad \text{oxidation state As = -3}\]

The oxidation number for \(Zn_{(s)} \) is 0 on the reactants side and +2 on the products side for aqueous Zn. From the half reactions we can determine that \(4Zn_{(s)} \) is oxidized as it loses electrons and gains a positive charge (i.e. 0 to +2) while As is reduced as it gains electrons and acquires a negative charge (i.e. +5 to -3). Now, we can conclude:

\( \textbf{Oxidant: } H_3AsO_{4(aq)} \qquad \textbf{Reductant: } Zn_{(s)} \)

\(\textbf{4).}\) For reaction 4 we see that the overall reaction \( 2NO_{2(g)} + 2OH^{ -}_{(aq)} \to NO^{ -}_{2(aq)} + NO^{-}_{3(aq)} + H_2O_{(l)}\) can be broken down into the following half reactions:

\[ NO_{2(g)} \to NO^{ -}_{2(aq)} \qquad NO_{2(g)} \to NO^{-}_{3(aq)}\]

We see that the oxidation number for N in gaseous \(NO_2 \) is +4 on the reactants side, but +3 on the products side for \( NO^{ -}_{2(aq)} \). Similarly, the oxidation number for N in gaseous \(NO_2 \) is +4 on the reactants side, but +5 on the products side for aqueous \(NO^{-}_3\). From the half reactions we can determine that \(NO_{2(g)} \) is both oxidized and reduced to form \(NO^{ -}_{2(aq)}\) and \(NO^{-}_{3(aq)} \). Now, we can conclude:

\( \textbf{Oxidant: } NO_{2(g)} \qquad \textbf{Reductant: } NO_{2(g)} \)

(reaction 4 is the most difficult since NO₂ acts as both a reductant and an oxidant. The student does a good job of explaining this and the OH^- is merely an indicator that it is in a basic solution.)-Yvanna Todorova

Q19.1.19

Predict the products of the following reactions and balance the equations.

1. Zn is added to a solution of Cr₂(SO₄)₃ in acid.
2. FeCl₂ is added to a solution containing an excess of Cr₂O₇²⁻ in hydrochloric acid.
3. Cr²⁺ is added to Cr₂O₇²⁻ in acid solution.
4. Mn is heated with CrO₃.
5. CrO is added to 2HNO₃ in water.
6. FeCl₃ is added to an aqueous solution of NaOH.

S19.1.19

\(\text{1).}\) In the first reaction, we have solid zinc being added to an acidic solution of \( Cr_2\text{(SO}_4\text{)}_3\) where we should see a single displacement reaction occur. The zinc atoms will replace the two chromium atoms in \( Cr_2\text{(SO}_4\text{)}_3\) to form \( Zn_2\text{(SO}_4\text{)}_3\), a soluble compound in aqueous solution. Thus, the final balanced chemical equation will be:

\( Cr_2\text{(SO}_4\text{)}_3{\text{(aq)}} + 2Zn{\text{(s)}} + H_3O^+{\text{(aq)}} \to 2Zn^{2+}{\text{(aq)}} + H_3O^+{\text{(aq)}} + 2Cr^{2+}{\text{(aq)}} + 3SO_4^{2-}{\text{(aq)}} \)

\(\text{2).}\)In the second reaction, we have \(FeCl_2\) being added to an excess solution of \( {Cr_2O_7}^{2-}_{(aq)}\) in hydrochloric acid where we should see a reaction occur (To clarify, hydrocholoric acid si strong which means it dissociates and thus it contributes a H⁺ and a Cl^-)-Yvanna Todorova. The iron ions will react with the chloride ions in solution to form \( {FeCl_3}_{\text{(aq)}}\) along with \( {CrCl_3}_{\text{(aq)}}\). Thus, the final balanced chemical equation will be:

\( {FeCl}_{2(aq)} + {Cr_2O_7}^{2-}_{(aq)} + {HCl}_{\text{(aq)}} \to {CrCl}_{3(aq)} + {H_2O}_{\text{(aq)}} + {FeCl_3}_{\text{(aq)}} \)

\(\text{3).}\)In the third reaction, we have Chromium (II) ion being added to an acidic solution of \( {Cr_2O_7}^{2-}_{(aq)}\) where we should see a reaction occur. The chromium ions will react with the hydronium ions in solution to form \( {Cr(OH)_2}_{\text{(s)}}\), an insoluble compound in aqueous solution. Thus, the final balanced chemical equation will be:

\( {Cr_2O_7}^{2-}_{(aq)} + Cr^{2+}_{\text{(s)}} + 2{H_3O}^+_{\text{(aq)}} \to {Cr_2O_7}^{2-}_{(aq)} + 4H^+_{\text{(aq)}} + {Cr(OH)_2}_{\text{(s)}} \)

\(\text{4).}\) In this reaction, we have solid manganese being heated with \( {CrO_3}_{(s)}\), called chromium trioxide, where we should see a reaction occur. The manganese solid will react with some of the chromate molecules to form \( 3{Mn_3O_4}_{\text{(s)}}\), called Manganese (II, III) oxide. The remaining chromium trioxide molecules are turned into chromium (II) oxide. Thus, the final balanced chemical equation will be:

\( 8{CrO_3}_{(s)} + 9Mn_{\text{(s)}} \overset{\Delta}{\to} 4{Cr_2O_3}_{(s)} + {Mn_3O_4}_{\text{(s)}} \)

\(\text{5).}\) In this reaction, we have solid chromium (II) oxide being added to \( 2{HNO_3}^-_{(aq)}\), where we should see a reaction occur. The chromium compound will lose an oxygen to form water molecules and the chromium (II) ion. Thus, the final balanced chemical equation will be:

\( {CrO}_{(s)} + {2H_3O}_{\text{(aq)}} + {2HNO_3}_{\text{(aq)}} \to {3H_2O}_{\text{(aq)}} + {2NO_3}^-_{\text{(aq)}} +Cr^{2+}_{\text{(aq)} }\)

\(\text{6).}\) In this reaction, we have solid iron (III) chloride being added to \( 3{NaOH}_{(aq)}\), where we should see a reaction occur(Connecting this to previous chem classes, Compunds with NO₃ are very soluble, this means that in reaction like these it is likely that it will dissolve and interact with other atoms)-Yvanna Todorova. The iron will react with hydroxide ions to form an insoluble metal hydroxide and other ions in solution. Thus, the final balanced chemical equation will be:

\( {FeCl_3}_{(s)} + 3{NaOH}_{\text{(aq)}} \to {Fe(OH)_3}_{\text{(s)}} + {3Na}^+_{\text{(aq)}} + 3Cl^-_{\text{(aq)}} \)

Q20.7.5

This reaction is characteristic of a lead storage battery: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

If you have a battery with an electrolyte that has a density of 1.15 g/cm³ and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?

S20.7.5

\[ \textbf{Initial Conditions: } \text{Density = } \frac{1.15g}{cm^3} \qquad Mass_{H_2SO_4} = \text{0.30} \bullet \text{ Mass}_{Total} \qquad \text{ Molecular mass }H_2SO_4 = \text{ 98.079 g}\]

We know that at standard conditions the cell will have concentrations of 1 M at 1 bar and 298 K. In this case we see that this cell is not in standard conditions and thus, it will have a different \( E_{cell}\) value than it would at standard conditions. First, we calculate the concentration of \( H_2SO_4\) using the density and mass percent. (It's important to be careful with units when converting and also it is best to use as many significant figures as possible when calculating so that the answer is precise)-Yvanna Todorova

\[ \text{Density = } \frac{mass}{volume} \implies \text{Mass = volume } \bullet \text{ density} \]

\[ \text{Mass}_{Total} \text{ = } (1 L)\frac{1000 mL}{L} \bullet \frac{1.15g}{mL} = \text{ 1150 g}\]

\[ Mass_{H_2SO_4} = \text{0.30} \bullet 1150 g = \text{ 345 g} \]

Now that we have the total mass of sulfuric acid in the cell, we can determine the number of moles of \( H_2SO_4\) using the molecular mass of sulfuric acid. This value is then used to determine the concentration of sulfuric acid in the cell.

\[ moles_{H_2SO_4} = \frac{mass_{H_2SO_4}}{\text{molar mass}_{H_2SO_4}}\]

\[ moles_{H_2SO_4} = \frac{345 g}{98.079 g} = \text{3.5176 mol H}_2SO_4 \]

\[ [H_2SO_4] = \frac{\text{mol H}_2SO_4}{1 L} = 3.52 M\]

After calculating the concentration of \( H_2SO_4\) we can determine the value of \( E_{cell}\) with regards to \( E^{\circ}_{cell}\) using the Nernst Equation. We can see from the balanced equation that the oxidation number of Pb in \( {PbO_2}_{(s)} \) is +4 and changes to +2 in the product \( {PbSO_4}_{(s)} \), indicating that there are 2 moles of \( e^-\) transferred in the reaction.

\[ E = E^{\circ} - \frac{0.0592V}{n} logQ \qquad @ Standard Conditions \]

\[ Q = \frac{1}{{[H_2SO_4]}^2} = \frac{1}{(3.52M)^2} \]

\[ E = E^{\circ} - \frac{0.0592V}{\text{2 mol {e}}^-} log \frac{1}{(3.52M)^2} \]

\[ E = E^{\circ} - (- 0.032208) \qquad \implies E > E^{\circ}\]

We can see that the value of \( E_{cell} \) is greater than the value of \( E^{\circ}_{cell}\).