Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

The Nernst equation relates cell potential under non-standard conditions to the reaction quotient, providing a direct method to determine unknown stoichiometric coefficients. By identifying the cathode and anode based on standard reduction potentials, the standard cell potential is calculated as $E^\ominus_{\text{cell}} = 0.15 - (-0.036) = 0.186 \text{ V}$.

The balanced cell reaction relies on equalizing the electrons transferred between the reduction of $\mathrm{M}^{x+}$ and the oxidation of $\mathrm{Fe}$, where the electron exchange $n$ becomes $3x$. Substituting the experimental values into the Nernst equation gives the following relationship:
$E_{\text{cell}} = E^\ominus_{\text{cell}} - \frac{0.059}{n} \log Q$
$0.2057 = 0.186 - \frac{0.059}{3x} \log(10^{-2})$
Simplifying the expression by finding the difference in potential results in $0.2057 - 0.186 = 0.0197$, while the logarithmic term $-\frac{0.059}{3x}(-2)$ simplifies to $\frac{0.118}{3x}$. Equating these terms yields $0.0197 = \frac{0.118}{3x}$, which simplifies to $3x = \frac{0.118}{0.0197} \approx 6$, confirming that the value of $x$ is 2.

According to the Nernst equation $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$, the cell potential increases when the reaction quotient Q = \frac{[\mathrm{Fe}^{2+}]\}{[\mathrm{Fe}^{3+}]$$[\mathrm{Cl}^-]$}$decreases for the net cell reaction$\mathrm{Ag}(\mathrm{s}) + \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{Cl}^-(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s}) + \mathrm{Fe}^{2+}(\mathrm{aq})$. Decreasing the value of$Q$requires reducing the concentration of the product$\mathrm{Fe}^{2+}$, raising the concentration of the reactant$\mathrm{Fe}^{3+}$, or raising the concentration of the reactant$\mathrm{Cl}^-$. Consequently, the cell potential increases under the conditions specified in options C, D, and E.

The strength of a reducing agent is determined by its oxidation potential, $E^{0}_{\text{ox}}$, which is the negative of the provided standard reduction potential, $E^{0}_{\text{red}}$. Comparing the oxidation potentials for the species involved, we find $\text{Al} \rightarrow \text{Al}^{3+} + 3e^{-}$ with $E^{0}_{\text{ox}} = 1.66 \, \text{V}$, $\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^{-}$ with $E^{0}_{\text{ox}} = 0.74 \, \text{V}$, $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-}$ with $E^{0}_{\text{ox}} = -0.77 \, \text{V}$, and $\text{Co}^{2+} \rightarrow \text{Co}^{3+} + e^{-}$ with $E^{0}_{\text{ox}} = -1.81 \, \text{V}$. Since a more positive oxidation potential indicates a stronger reducing agent, the tendency to act as a reducing agent follows the decreasing order $\text{Al} > \text{Cr} > \text{Fe}^{2+} > \text{Co}^{2+}$.

The molar conductivity of a weak acid is given by $\Lambda_m = \alpha \Lambda_m^{\circ}$, where the degree of dissociation is $\alpha = \sqrt{K_a / C}$ for weak electrolytes with $\alpha \ll 1$. Given $\lambda^{\circ}_{Q^-} = \lambda^{\circ}_{Z^-}$, the limiting molar conductivities $\Lambda_m^{\circ}(HQ) = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{Q^-}$ and $\Lambda_m^{\circ}(HZ) = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{Z^-}$ are identical. The ratio of the molar conductivities is therefore $\frac{\Lambda_m(HQ)}{\Lambda_m(HZ)} = \sqrt{\frac{K_a(HQ)}{C_1}} \cdot \sqrt{\frac{C_2}{K_a(HZ)}} = \frac{1}{30}$. Substituting $C_1 = 0.18 \text{ M}$ and $C_2 = 0.02 \text{ M}$ into this relation yields $\sqrt{\frac{K_a(HQ)}{K_a(HZ)}} \cdot \sqrt{\frac{0.02}{0.18}} = \frac{1}{30}$, which simplifies to $\sqrt{\frac{K_a(HQ)}{K_a(HZ)}} \cdot \frac{1}{3} = \frac{1}{30}$ or $\sqrt{\frac{K_a(HQ)}{K_a(HZ)}} = \frac{1}{10}$. Squaring both sides gives the ratio $\frac{K_a(HQ)}{K_a(HZ)} = 10^{-2}$. Since $pK_a = -\log K_a$, the difference is $pK_a(HQ) - pK_a(HZ) = -\log(K_a(HQ)) + \log(K_a(HZ)) = -\log\left(\frac{K_a(HQ)}{K_a(HZ)}\right) = -\log(10^{-2}) = 2$.

The standard reduction potential for the metal/metal insoluble salt electrode, $\mathrm{MX}(s) + e^- \rightarrow \mathrm{M}(s) + \mathrm{X}^-(aq)$, is derived from the net free energy change of the reduction $\mathrm{M}^+(aq) + e^- \rightarrow \mathrm{M}(s)$ combined with the solubility equilibrium $\mathrm{MX}(s) \rightleftharpoons \mathrm{M}^+(aq) + \mathrm{X}^-(aq)$. The relation between the standard potential, the metal ion reduction potential, and the solubility constant is:
$E^\ominus_{\mathrm{X}^-/\mathrm{MX}/\mathrm{M}} = E^\ominus_{\mathrm{M}^+/\mathrm{M}} + \frac{0.059}{n} \log_{10} K_{sp}$
Using $E^\ominus_{\mathrm{M}^+/\mathrm{M}} = 0.79\,\text{V}$, $n=1$, and $K_{sp} = 10^{-10}$:
$E^\ominus = 0.79\,\text{V} + 0.059\,\text{V} \times \log_{10}(10^{-10})$
$E^\ominus = 0.79\,\text{V} - 0.59\,\text{V} = 0.20\,\text{V}$
The final value in millivolts is 200 mV.

The conductivity of a solution is determined by its conductance $G$ and the cell constant $\frac{l}{A}$ using the relation $\kappa = G \times \frac{l}{A}$. Given the conductance $G = 4 \times 10^{-5}$ S, the distance between electrode plates $l = 15$ cm, and the surface area $A = 1$ cm², the cell constant is $15$ cm$^{-1}$, which is equivalent to $1500$ m$^{-1}$ in SI units. Substituting these values into the equation gives $\kappa = (4 \times 10^{-5} \text{ S}) \times (1500 \text{ m}^{-1}) = 0.06$ S m$^{-1}$.

For a weak acid $HX$ that dissociates according to $HX \rightleftharpoons H^+ + X^-$, the conductivity is expressed as \kappa = \[H^+]$ \Lambda_m^\circ$, where$[H^+]$is the concentration of hydrogen ions and$\Lambda_m^\circ$is the limiting molar conductivity. The pH value of 5 corresponds to a hydrogen ion concentration of$[H^+] = 10^{-5}$M, which converts to$10^{-2}$mol m$^{-3}$in SI units. Substituting the calculated conductivity and concentration into the equation yields$\Lambda_m^\circ = \frac{\kappa}{[H^+]$} = \frac{0.06}{10^{-2}}$, resulting \in a value of$6$S m$^2$mol$^{-1}$.

The overall electrochemical cell reaction is obtained by combining the anodic oxidation of $\mathrm{HSnO_2^-}$ and the cathodic reduction of $\mathrm{Bi_2O_3}$, which is $3\mathrm{HSnO_2^-} + \mathrm{Bi_2O_3(s)} + 3\mathrm{H_2O} + 3\mathrm{OH^-} \rightarrow 3\mathrm{Sn(OH)_6^{2-}} + 2\mathrm{Bi(s)}$. This reaction involves the transfer of $n=6$ electrons. The standard cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.44\text{ V} - (-0.9\text{ V}) = 0.46\text{ V}$.

Applying the Nernst equation to this cell with the given concentrations:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{6} \log \frac{[\mathrm{Sn(OH)_6^{2-}}]^3}{[\mathrm{HSnO_2^-}]^3 \$[\mathrm{OH^-}]^3}$
$0.2353 = 0.46 - \frac{0.059}{2} \log \frac{0.5}{0.05 \$[\mathrm{OH^-}]\$}$
$0.2353 = 0.46 - 0.0295 \log \left( \frac{10}{[\mathrm{OH^-}]\$} \right)$
$0.2353 = 0.46 - 0.0295(1 - \log \$[\mathrm{OH^-}]\$)$
$0.2353 = 0.4305 + 0.0295 \log [\mathrm{OH^-}]$
Solving for the concentration of hydroxide ions yields \log \[\mathrm{OH^-}]$ = -6.6$, which corresponds to a$\mathrm{pOH}$of$6.6$. The$\mathrm{pH}$of the buffer is then$14 - 6.6 = 7.4$.

Using the Henderson-Hasselbalch equation for the bicarbonate buffer:
$\mathrm{pH} = \mathrm{pK_a} + \log \frac{[\mathrm{HCO_3^-}]\$}{[\mathrm{H_2CO_3}]}$
$7.4 = 6.11 + \log \left( \frac{5x}{20} \right)$
$1.29 = \log \left( \frac{x}{4} \right)$
$\frac{x}{4} = \text{antilog}(1.29) = 19.5$
$x = 78$

The core principle for spontaneous electrochemical cells is that the half-cell with the higher standard reduction potential acts as the cathode. Given $\mathrm{AgBr}(\mathrm{s})$ with $E^0 = +0.07 \mathrm{V}$ and $\mathrm{Fe}(\mathrm{OH})_2(\mathrm{s})$ with $E^0 = -0.88 \mathrm{V}$, the silver-bromide reaction serves as the reduction half-reaction. The standard cell potential is calculated as $E^0_{\text{cell}} = +0.07 \mathrm{V} - (-0.88 \mathrm{V}) = +0.95 \mathrm{V}$.

To determine the overall cell reaction, the iron half-reaction must be reversed to represent oxidation: $\mathrm{Fe}(\mathrm{s}) + 2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Fe}(\mathrm{OH})_2(\mathrm{s}) + 2 e^{-}$. When this is added to the reduction of two moles of $\mathrm{AgBr}(\mathrm{s})$, the total spontaneous reaction becomes $\mathrm{Fe}(\mathrm{s}) + 2 \mathrm{OH}^{-}(\mathrm{aq}) + 2 \mathrm{AgBr}(\mathrm{s}) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_2(\mathrm{s}) + 2 \mathrm{Ag}(\mathrm{s}) + 2 \mathrm{Br}^{-}(\mathrm{aq})$. Because iron is oxidized, it is not reduced, and the cell potential remains an intensive property that does not depend on the stoichiometric coefficients of the balanced equation.

In any electrochemical cell, the overall redox reaction is obtained by balancing the half-reactions at the anode and cathode to ensure that the total number of electrons lost equals the number gained. Multiplying the anode half-reaction by $n_2$ and the cathode half-reaction by $n_1$ ensures that the electrons are consumed exactly as they are produced, resulting in the following balanced process:
$n_2 \mathrm{Red}_1 + n_1 \mathrm{Ox}_2 \rightarrow n_2 \mathrm{Ox}_1^{n_1^{+}} + n_1 \mathrm{Red}_2^{n_2^{-}}$
In this context, $n$ refers to the number of electrons transferred in the redox reaction, and because electrical work is physically defined as the product of charge and potential difference, describing it as the ratio of these two quantities is incorrect.

The relationship between the terminal potential difference ($V$) and the current ($I$) in a closed circuit Daniell cell is described by the equation $V = E_{cell} - Ir$, where $E_{cell}$ is the electromotive force and $r$ is the internal resistance. This linear equation indicates that as the current drawn from the cell increases, the terminal voltage decreases linearly with a constant negative slope equal to $-r$. Plot D accurately depicts this behavior, showing a straight line starting from the EMF on the vertical axis and dropping as current increases.

The equation $\Lambda_m = \Lambda_m^{\circ} - A c^{1/2}$ describes the linear relationship between molar conductivity and the square root of the concentration for strong electrolytes, where the slope of the line is equal to $-A$. To determine the value of $A$, first calculate the square root of the given concentration values: $\sqrt{0.04} = 0.2$, $\sqrt{0.09} = 0.3$, $\sqrt{0.16} = 0.4$, and $\sqrt{0.25} = 0.5$. Using the data points $(0.2, 96.1)$ and $(0.3, 95.7)$, the slope is calculated as the change in $\Lambda_m$ divided by the change in $c^{1/2}$, resulting in $\frac{95.7 - 96.1}{0.3 - 0.2} = \frac{-0.4}{0.1} = -4$. Because the slope corresponds to the coefficient $-A$ in the linear equation, $-A = -4$, which gives $A = 4$.

For a concentration cell involving the same metal electrodes, the cell potential is determined by the Nernst equation $E_{cell} = -\frac{0.0591}{n} \log \left( \frac{[M^+]_{anode}}{[M^+]_{cathode}} \right)$, where the standard cell potential is zero. To obtain a positive cell potential ($E_{cell} > 0$), the logarithmic term must be negative, which requires the concentration of the electrolyte at the cathode to be greater than the concentration at the anode. Consequently, assigning $c_1$ to the cathode such that $c_1 > c_2$ ensures [M^+]_{cathode} > \[M^+]_{anode}$, resulting in a spontaneous reaction and a positive cell potential.

To determine the concentration of silver ions in the voltaic cell, we apply the Nernst equation, which links the cell potential to standard conditions and the reaction quotient. We first calculate the standard cell potential by taking the difference between the reduction potentials of the cathode and anode, resulting in $0.80\,\mathrm{V} - (-0.76\,\mathrm{V}) = 1.56\,\mathrm{V}$. The spontaneous reaction involves zinc being oxidized and silver ions being reduced, represented by the equation $\mathrm{Zn} + 2\mathrm{Ag}^{+} \longrightarrow \mathrm{Zn}^{2+} + 2\mathrm{Ag}$, which clearly shows that two electrons are exchanged, so $n = 2$.

The Nernst equation is expressed as $E_{\mathrm{cell}} = E^{\Theta}_{\mathrm{cell}} - \frac{0.059}{n} \log Q$, where the reaction quotient $Q$ is defined by the ratio of ion concentrations, \frac{[\mathrm{Zn}^{2+}]\}{[\mathrm{Ag}^{+}]^2}$. Given that the zinc ion concentration is$1,\mathrm{M}$and the silver ion concentration is$x$, the quotient simplifies to$\frac{1}{x^2}$. Substituting the known potentials and the value of$n$into the equation gives$1.60 = 1.56 - \frac{0.059}{2} \log \left( \frac{1}{x^2} \right)$. Because$\log \left( \frac{1}{x^2} \right)$is equivalent to$-2 \log x$, the expression becomes$1.60 = 1.56 + \frac{0.059}{2} × 2 \log x$. This simplifies to$1.60 = 1.56 + 0.059 \log x$, and rearranging the terms isolates$0.04 = 0.059 \log x$, which gives$\log x = \frac{4}{5.9}$.

Since standard reduction potential is an intensive property, it cannot be added or subtracted directly; instead, you must relate the reactions using the extensive property of Gibbs free energy, defined by $\Delta G^{\ominus} = -nFE^{\ominus}$, where $n$ represents the number of electrons transferred and $F$ is Faraday's constant. To find the potential for the reaction $\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}$, identify the associated half-reactions: the reduction of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}$ has a potential $X$ and involves $n=2$, giving $\Delta G^{\ominus}_{1} = -2FX$, while the reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}$ has a potential $Y$ and involves $n=3$, giving $\Delta G^{\ominus}_{2} = -3FY$. The target reaction is obtained by subtracting the first half-reaction from the second, which implies the relationship $\Delta G^{\ominus}_{3} = \Delta G^{\ominus}_{2} - \Delta G^{\ominus}_{1}$. Substituting the free energy expressions into this equation results in $-1F E^{\ominus}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = -3FY - (-2FX)$, which simplifies to $-F E^{\ominus}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = -3FY + 2FX$. Dividing both sides by $-F$ isolates the required potential as $3Y - 2X$.

The electrochemical cell potential ($E_{\text{cell}}$) must be positive for the spontaneous evolution of oxygen gas at the anode, where the oxidation reaction is $2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H^+} + 4\mathrm{e^-}$. The cell consists of an oxygen anode and a metal cathode ($\mathrm{M^{2+}}/\mathrm{M}$), with the standard reduction potential of the oxygen electrode being $E_{\mathrm{O_2/H_2O}}^{\circ} = 1.23\,\mathrm{V}$ and that of the metal electrode being $E_{\mathrm{M^{2+}/M}}^{\circ} = 0.994\,\mathrm{V}$.

The Nernst equation for the oxygen electrode acting as an anode is expressed in terms of its reduction potential, which is given by $E_{\mathrm{red}} = E_{\mathrm{O_2/H_2O}}^{\circ} - \frac{0.0591}{4} \log \frac{1}{[\mathrm{H^+}]^4} = 1.23 - 0.0591 \mathrm{pH}$. Since the oxygen electrode is the anode, its oxidation potential is the negative of this reduction potential, $E_{\mathrm{anode}} = -(1.23 - 0.0591 \mathrm{pH}) = -1.23 + 0.0591 \mathrm{pH}$. The cell potential is the sum of the cathode reduction potential and the anode oxidation potential, resulting in:

$E_{\text{cell}} = E_{\mathrm{M^{2+}/M}}^{\circ} + E_{\text{anode}} = 0.994 + (-1.23 + 0.0591 \mathrm{pH})$

Setting the condition for spontaneous oxygen evolution at the anode as $E_{\text{cell}} > 0$:

$0.994 - 1.23 + 0.0591 \mathrm{pH} > 0$
$-0.236 + 0.0591 \mathrm{pH} > 0$
$\mathrm{pH} > \frac{0.236}{0.0591} \approx 3.99$

Thus, the pH above which oxygen gas would start to evolve at the anode is approximately 4.

According to Faraday's laws of electrolysis, the total charge passed through the solution is proportional to the number of electrons transferred, which determines the amount of substance produced at the electrodes. The deposition of 300 mg of copper from $\mathrm{Cu}^{2+}$ requires $n_1 = 2 \times \frac{0.3}{63.54} \approx 0.009443$ moles of electrons, corresponding to a charge $Q_1 = 0.009443 \times 96500 \approx 911.27$ C. In the subsequent phase, a current of 600 mA (0.6 A) passed for 28 minutes (1680 seconds) provides an additional charge $Q_2 = 0.6 \times 1680 = 1008$ C.

The total charge passed through the system during both phases is $Q_{total} = 911.27 + 1008 = 1919.27$ C, which corresponds to $n_{e} = \frac{1919.27}{96500} \approx 0.019889$ moles of electrons. At the anode, the evolution of oxygen follows the reaction $2\mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{O}_2 + 4\mathrm{H}^{+} + 4e^{-}$, where 1 mole of $\mathrm{O}_2$ is produced for every 4 moles of electrons. Consequently, the total moles of oxygen produced is $n_{\mathrm{O}_2} = \frac{0.019889}{4} \approx 0.004972$ moles. Given the molar volume at STP is 22400 mL/mol, the volume of oxygen evolved is $V = 0.004972 \times 22400 \approx 111.37$ mL, which rounds to 111 mL.

Molar conductivity provides a direct link between the electrolytic conductivity of a solution and its concentration, expressed by the relationship $\Lambda_m = \frac{\kappa \times 1000}{M}$, where $\kappa$ denotes specific conductivity and $M$ is the molarity. The specific conductivity is determined by multiplying the measured conductance $G$ by the cell constant $G^*$. Using the given values $G = 1.9 \times 10^{-3} \text{ S}$ and $G^* = 1.3 \text{ cm}^{-1}$ alongside the provided molar conductivity of $123.5 \text{ S cm}^2 \text{mol}^{-1}$, the molarity of the MX solution is determined to be 2 mol L$^{-1}$.

The weight percentage of the solute is calculated by comparing the mass of the solute to the total mass of the solution. In 1 L of this aqueous solution, the total mass is $1000 \text{ g}$ based on the given density of $1.0 \text{ g mL}^{-1}$. Since the molarity is 2 mol L$^{-1}$, the solution contains 2 moles of the solute. With the molar mass of MX being $75 \text{ g mol}^{-1}$, the total mass of the solute is $2 \times 75 = 150 \text{ g}$. The weight percentage is found by dividing the mass of the solute by the mass of the solution and multiplying by 100, which results in 15.

The energy efficiency of a fuel cell is determined by the relationship between the Gibbs free energy change and the standard cell potential, defined by the equation $\Delta G = -nFE^{\circ}_{\text{cell}}$. To calculate this, one must first identify the anodic and cathodic processes. By treating the methanol half-reaction as the oxidation step and reversing its reduction potential, the oxidation potential is found to be $-0.02 \, \mathrm{V}$. Combining this with the cathodic oxygen reduction potential of $1.23 \, \mathrm{V}$ results in a standard cell potential of $E^{\circ}_{\text{cell}} = 1.23 \, \mathrm{V} - 0.02 \, \mathrm{V} = 1.21 \, \mathrm{V}$.

With $n = 6$ electrons involved in the balanced redox reaction, the theoretical maximum work output is calculated as $\Delta G = -6 \times 96500 \times 1.21 = -700590 \, \mathrm{J}$. Because the fuel cell operates at an efficiency of $80\%$, the actual electrical work delivered is $0.8 \times 700590 \, \mathrm{J}$, which equals $560472 \, \mathrm{J}$. Expressing this quantity in units of kilojoules provides the final result of 560.

The value of $n$ in the Nernst equation corresponds to the number of moles of electrons transferred in the balanced redox reaction. The oxidation half-reaction for borohydride is $\mathrm{BH}_4^{-} + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_2\mathrm{BO}_3^{-} + 8\mathrm{H}^{+} + 8\mathrm{e}^{-}$, while the reduction half-reaction for the chlorate ion is $\mathrm{ClO}_3^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}$. To balance the electron transfer between the two half-reactions, the oxidation process must be multiplied by 3 and the reduction process by 4, resulting in a common transfer of $3 \times 8 = 24$ and $4 \times 6 = 24$ electrons. Therefore, the total number of electrons involved in the balanced reaction is $n = 24$.

The Nernst equation, $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q$, provides the cell potential based on standard conditions and the reaction quotient. The standard cell potential is $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = -0.44 \, \mathrm{V} - (-0.90 \, \mathrm{V}) = 0.46 \, \mathrm{V}$. The oxidation half-reaction is $3(\mathrm{HSnO}_2^{-} + 4\mathrm{OH}^{-} \rightarrow \mathrm{Sn}(\mathrm{OH})_6^{2-} + 2e^{-})$ and the reduction half-reaction is $\mathrm{Bi}_2\mathrm{O}_3 + 3\mathrm{H}_2\mathrm{O} + 6e^{-} \rightarrow 2\mathrm{Bi} + 6\mathrm{OH}^{-}$, determining the number of electrons transferred as $n=6$. Substituting these values and $Q = 10^6$ into the Nernst equation yields $E_{\text{cell}} = 0.46 - \frac{0.059}{6} \log(10^6) = 0.46 - 0.059 = 0.401 \, \mathrm{V}$. As this is $4.01 \times 10^{-1} \, \mathrm{V}$, the nearest integer is 4.