An aqueous solution contains 1.0 M of $\mathrm{X}^{2+}$ and 0.001 M of $\mathrm{Y}^2+$ ions at $25^{...

Given Reduction Potentials:$$ E_{X^{2+} / X}^0=0.35 \mathrm{~V} $$$$ E_{Y^{2+} / Y}^0=0.40 \mathrm{~V} $$The reduction potential for $Y^{2+}$ is higher than for $X^{2+}$. This means that, under standard conditions, $Y^{2+}$ would be reduced first. However, the significant difference in concentration affects the actual reduction process.Nernst Equation:The Nernst equation helps us account for the non-standard conditions$$ E=E^0-\frac{0.0591}{n} \log Q $$For $X^{2+}$ :$$ E_{X^{2+} / X}=0.35-\frac{0.0591}{2} \log \frac{1}{1.0}=0.35 \mathrm{~V} $$For $Y^{2+}$ :$$ E_{Y^{2+} / Y}=0.40-\frac{0.0591}{2} \log \frac{1}{0.001}=0.40-0.0886=0.3114 \mathrm{~V} $$Comparing Effective Potentials:Effective potential for $X^{2+}$ is 0.35 V .Effective potential for $Y^{2+}$ is 0.3114 V .The material deposited on the cathode will be only $X$ due to its higher effective potential despite the higher standard reduction potential of $Y^{2+}$

Consider the following data for KCl solution at a particular temperature. What is the value of the l...

Given data:$$ \begin{aligned} & C=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \rightarrow \Lambda_m=149.1 \mathrm{Scm}^2 / \mathrm{mol} \\ & C=9 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \rightarrow \Lambda_m=147.1 \mathrm{Scm}^2 / \mathrm{mol} \end{aligned} $$Using the formula for extrapolation:$$ \Lambda_m=\Lambda_m^0-K \sqrt{C} $$(where $\Lambda_m^0$ is the limiting molar conductivity, $K$ is a constant, and $C$ is the concentration.)We have two equations:$$ \begin{aligned} & 149.1=\Lambda_m^0-K \sqrt{1 \times 10^{-4}} \\ & 147.1=\Lambda_m^0-K \sqrt{9 \times 10^{-4}} \end{aligned} $$Subtracting:$$ \begin{gathered} 149.1-147.1=\left(\Lambda_m^0-0.01 K\right)-\left(\Lambda_m^0-0.03 K\right) \\ 2=0.02 K \\ K=100 \end{gathered} $$Now, substitute $K=100$ back into one of the original equations to find $\Lambda_m^0$$$ \begin{gathered} 149.1=\Lambda_m^0-0.01 \times 100 \\ 149.1=\Lambda_m^0-1 \\ \Lambda_m^0=149.1+1=150.1 \mathrm{Sem}^2 / \mathrm{mol} \end{gathered} $$So, the value of the limiting molar conductivity $\Lambda_m^0$ for KCl is $150.1 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol}$.

During the charging and discharging of a lead-acid battery (a Pb anode, a grid of Pb packed with $\m...

Discharging Process (Battery operating as a galvanic cell):At the Anode (Negative electrode):Lead (Pb, oxidation state 0) is oxidized to Lead(II) sulfate (PbSO₄, oxidation state +2).$\mathrm{Pb(s)} + \mathrm{SO_4^{2-}(aq)} \rightarrow \mathrm{PbSO_4(s)} + 2\mathrm{e^-}$This corresponds to the elemental reaction: $\mathrm{Pb} \rightarrow \mathrm{Pb}^{2+}+2 \mathrm{e}^{-}$ (if we consider soluble Pb²⁺, though it immediately precipitates as PbSO₄). This is one of the options (Option C).At the Cathode (Positive electrode):Lead(IV) dioxide (PbO₂, oxidation state +4) is reduced to Lead(II) sulfate (PbSO₄, oxidation state +2).$\mathrm{PbO_2(s)} + \mathrm{SO_4^{2-}(aq)} + 4\mathrm{H^+(aq)} + 2\mathrm{e^-} \rightarrow \mathrm{PbSO_4(s)} + 2\mathrm{H_2O(l)}$This corresponds to the elemental change: $\mathrm{Pb}^{4+} + 2\mathrm{e}^- \rightarrow \mathrm{Pb}^{2+}$ (considering solvated ions for simplicity in states).Charging Process (Battery operating as an electrolytic cell):This is the reverse of the discharging process.At the Anode (Positive electrode during charging):Lead(II) sulfate (PbSO₄, oxidation state +2) is oxidized back to Lead(IV) dioxide (PbO₂, oxidation state +4).$\mathrm{PbSO_4(s)} + 2\mathrm{H_2O(l)} \rightarrow \mathrm{PbO_2(s)} + \mathrm{SO_4^{2-}(aq)} + 4\mathrm{H^+(aq)} + 2\mathrm{e^-}$This corresponds to the elemental reaction: $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$ (Option B).At the Cathode (Negative electrode during charging):Lead(II) sulfate (PbSO₄, oxidation state +2) is reduced back to Lead (Pb, oxidation state 0).$\mathrm{PbSO_4(s)} + 2\mathrm{e^-} \rightarrow \mathrm{Pb(s)} + \mathrm{SO_4^{2-}(aq)}$This corresponds to the elemental reaction: $\mathrm{Pb}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Pb}$.Overall transformation during charging (Option D):The overall reaction during charging is:$2\mathrm{PbSO_4(s)} + 2\mathrm{H_2O(l)} \rightarrow \mathrm{Pb(s)} + \mathrm{PbO_2(s)} + 2\mathrm{H_2SO_4(aq)}$In terms of lead oxidation states, two moles of Pb(II) (from PbSO₄) transform into one mole of Pb(0) and one mole of Pb(IV) (from PbO₂). This is a disproportionation reaction for lead(II): one Pb(II) is reduced to Pb(0), and another Pb(II) is oxidized to Pb(IV).This exactly matches Option D: $2 \mathrm{~Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}+\mathrm{Pb}$.Let's examine each option:Option A: $\mathrm{Pb}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$This reaction represents the direct reduction of Lead(IV) to elemental Lead(0). While Lead(IV) (in PbO₂) is reduced during discharge, it is reduced to Lead(II) (in PbSO₄), which is a 2-electron process. Elemental Lead(0) is formed from Lead(II) during charging, also a 2-electron process. There is no single, direct 4-electron reduction step from Pb(IV) to Pb(0) as a primary reaction in the lead-acid battery cycle.Option B: $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$This reaction occurs during charging at the positive electrode, where Pb(II) from PbSO₄ is oxidized to Pb(IV) in PbO₂. This reactiondoes occur.Option C: $\mathrm{Pb} \rightarrow \mathrm{Pb}^{2+}+2 \mathrm{e}^{-}$This reaction occurs during discharging at the negative electrode, where elemental Pb(0) is oxidized to Pb(II) in PbSO₄. This reactiondoes occur.Option D: $2 \mathrm{~Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}+\mathrm{Pb}$This reaction represents the overall transformation of lead(II) during charging, where it acts as both a reducing agent (to form Pb(0)) and an oxidizing agent (to form Pb(IV)). This is the net result of the two half-reactions during charging involving Pb(II) species, and thus it represents a process thatdoes occur.Therefore, the only reaction that does NOT occur as a primary direct step in the lead-acid battery mechanism is Option A.The final answer is $\boxed{A}$.

Electrochemistry - IAT-IISER Chemistry PYQs

Discharging Process (Battery operating as a galvanic cell):At the Anode (Negative electrode):Lead (Pb, oxidation state 0) is oxidized to Lead(II) sulfate (PbSO₄, oxidation state +2).

%5Cmathrm%7BPb(s)%7D%20%2B%20%5Cmathrm%7BSO_4%5E%7B2-%7D(aq)%7D%20%5Crightarrow%20%5Cmathrm%7BPbSO_4(s)%7D%20%2B%202%5Cmathrm%7Be%5E-%7D

This corresponds to the elemental reaction:

%5Cmathrm%7BPb%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B2%2B%7D%2B2%20%5Cmathrm%7Be%7D%5E%7B-%7D

(if we consider soluble Pb²⁺, though it immediately precipitates as PbSO₄). This is one of the options (Option C).At the Cathode (Positive electrode):Lead(IV) dioxide (PbO₂, oxidation state +4) is reduced to Lead(II) sulfate (PbSO₄, oxidation state +2).

%5Cmathrm%7BPbO_2(s)%7D%20%2B%20%5Cmathrm%7BSO_4%5E%7B2-%7D(aq)%7D%20%2B%204%5Cmathrm%7BH%5E%2B(aq)%7D%20%2B%202%5Cmathrm%7Be%5E-%7D%20%5Crightarrow%20%5Cmathrm%7BPbSO_4(s)%7D%20%2B%202%5Cmathrm%7BH_2O(l)%7D

This corresponds to the elemental change:

%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%20%2B%202%5Cmathrm%7Be%7D%5E-%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B2%2B%7D

(considering solvated ions for simplicity in states).Charging Process (Battery operating as an electrolytic cell):This is the reverse of the discharging process.At the Anode (Positive electrode during charging):Lead(II) sulfate (PbSO₄, oxidation state +2) is oxidized back to Lead(IV) dioxide (PbO₂, oxidation state +4).

%5Cmathrm%7BPbSO_4(s)%7D%20%2B%202%5Cmathrm%7BH_2O(l)%7D%20%5Crightarrow%20%5Cmathrm%7BPbO_2(s)%7D%20%2B%20%5Cmathrm%7BSO_4%5E%7B2-%7D(aq)%7D%20%2B%204%5Cmathrm%7BH%5E%2B(aq)%7D%20%2B%202%5Cmathrm%7Be%5E-%7D

This corresponds to the elemental reaction:

%5Cmathrm%7BPb%7D%5E%7B2%2B%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%2B2%20%5Cmathrm%7Be%7D%5E%7B-%7D

(Option B).At the Cathode (Negative electrode during charging):Lead(II) sulfate (PbSO₄, oxidation state +2) is reduced back to Lead (Pb, oxidation state 0).

%5Cmathrm%7BPbSO_4(s)%7D%20%2B%202%5Cmathrm%7Be%5E-%7D%20%5Crightarrow%20%5Cmathrm%7BPb(s)%7D%20%2B%20%5Cmathrm%7BSO_4%5E%7B2-%7D(aq)%7D

This corresponds to the elemental reaction:

%5Cmathrm%7BPb%7D%5E%7B2%2B%7D%20%2B%202%5Cmathrm%7Be%7D%5E-%20%5Crightarrow%20%5Cmathrm%7BPb%7D

.Overall transformation during charging (Option D):The overall reaction during charging is:

2%5Cmathrm%7BPbSO_4(s)%7D%20%2B%202%5Cmathrm%7BH_2O(l)%7D%20%5Crightarrow%20%5Cmathrm%7BPb(s)%7D%20%2B%20%5Cmathrm%7BPbO_2(s)%7D%20%2B%202%5Cmathrm%7BH_2SO_4(aq)%7D

In terms of lead oxidation states, two moles of Pb(II) (from PbSO₄) transform into one mole of Pb(0) and one mole of Pb(IV) (from PbO₂). This is a disproportionation reaction for lead(II): one Pb(II) is reduced to Pb(0), and another Pb(II) is oxidized to Pb(IV).This exactly matches Option D:

2%20%5Cmathrm%7B~Pb%7D%5E%7B2%2B%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%2B%5Cmathrm%7BPb%7D

.Let's examine each option:Option A:

%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%2B4%20%5Cmathrm%7Be%7D%5E%7B-%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D

This reaction represents the direct reduction of Lead(IV) to elemental Lead(0). While Lead(IV) (in PbO₂) is reduced during discharge, it is reduced to Lead(II) (in PbSO₄), which is a 2-electron process. Elemental Lead(0) is formed from Lead(II) during charging, also a 2-electron process. There is no single, direct 4-electron reduction step from Pb(IV) to Pb(0) as a primary reaction in the lead-acid battery cycle.Option B:

%5Cmathrm%7BPb%7D%5E%7B2%2B%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%2B2%20%5Cmathrm%7Be%7D%5E%7B-%7D

This reaction occurs during charging at the positive electrode, where Pb(II) from PbSO₄ is oxidized to Pb(IV) in PbO₂. This reactiondoes occur.Option C:

%5Cmathrm%7BPb%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B2%2B%7D%2B2%20%5Cmathrm%7Be%7D%5E%7B-%7D

This reaction occurs during discharging at the negative electrode, where elemental Pb(0) is oxidized to Pb(II) in PbSO₄. This reactiondoes occur.Option D:

2%20%5Cmathrm%7B~Pb%7D%5E%7B2%2B%7D%20%5Crightarrow%20%5Cmathrm%7BPb%7D%5E%7B4%2B%7D%2B%5Cmathrm%7BPb%7D

This reaction represents the overall transformation of lead(II) during charging, where it acts as both a reducing agent (to form Pb(0)) and an oxidizing agent (to form Pb(IV)). This is the net result of the two half-reactions during charging involving Pb(II) species, and thus it represents a process thatdoes occur.Therefore, the only reaction that does NOT occur as a primary direct step in the lead-acid battery mechanism is Option A.The final answer is

%5Cboxed%7BA%7D

Electrochemistry

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Electrochemistry

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