Electrochemical cells Questions in English

(A) An electrochemical cell functions as a galvanic cell when the external potential $(E_{ext})$ is less than the cell potential $(E_{cell})$,allowing spontaneous chemical reactions to produce electricity.
However,when an external potential $(E_{ext})$ greater than the cell potential $(E_{cell})$ is applied,the direction of the current is reversed,and the cell behaves as an electrolytic cell,where non-spontaneous reactions are driven by the external power source.
Therefore,the correct condition is $E_{ext} > E_{cell}$.

(A) In a mercury cell,the anode is made of zinc amalgam $(Zn(Hg))$ and the cathode is a paste of mercury$(II)$ oxide $(HgO)$ mixed with carbon $(C)$.
Therefore,the correct option is $A$.

(A) In a dry cell (Leclanché cell),the anode is made of zinc $(Zn)$ and the cathode is a carbon rod surrounded by powdered manganese dioxide $(MnO_2)$ and carbon.
During the discharge process,the $MnO_2$ acts as an oxidising agent where $Mn^{4+}$ is reduced to $Mn^{3+}$.
The reaction at the cathode is: $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$.
Therefore,$MnO_2$ is the oxidising agent.

(B) When an external opposing potential $(E_{ext})$ is applied to an electrochemical cell and increased slowly,the reaction continues to take place in the same direction until $E_{ext} = E_{cell}$.
If $E_{ext}$ is increased further such that $E_{ext} > E_{cell}$,the reaction reverses and the cell behaves as an electrolytic cell.

(A) The correct answer is $A$.
An electrolytic cell is a device that converts electrical energy into chemical energy,which is a non-spontaneous process.
In contrast,Leclanche cells,storage cells,and fuel cells are all types of electrochemical (galvanic/voltaic) cells that convert chemical energy into electrical energy through spontaneous redox reactions.

(B) In a mercury cell,the electrolyte is a paste of potassium hydroxide $(KOH)$ and zinc oxide $(ZnO)$.
Therefore,the correct mixture used as an electrolyte paste is $KOH + ZnO$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for a weak electrolyte like $NH_4OH$ can be calculated using strong electrolytes.
$\Lambda_{m(NH_4OH)}^0 = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$
To obtain this,we combine the molar conductivities of $NH_4Cl$,$NaOH$,and $NaCl$ as follows:
$\Lambda_{m(NH_4Cl)}^0 = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$
By performing the operation $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$,we get:
$(\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0 = \Lambda_{m(NH_4OH)}^0$
Therefore,the correct expression is $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$.

(C) In a Galvanic cell:
$(a)$ Electrons flow from anode to cathode,so conventional current flows from cathode to anode. This statement is correct.
$(b)$ The anode is the negative terminal where oxidation occurs. This statement is incorrect.
$(c)$ For a spontaneous reaction,$E_{cell} > 0$ and $\Delta G < 0$. If $E_{cell} < 0$,the reaction is non-spontaneous. This statement is incorrect.
$(d)$ The cathode is the positive terminal where reduction occurs. This statement is correct.
Therefore,statements $(a)$ and $(d)$ are correct.

A chemical reaction is possible when $AgNO_{3}$ solution is stirred with copper spoon as copper has lesser positive value of standard electrode reduction potential as compared to $Ag$.
The reaction will be,
$2 AgNO_{3} + Cu \longrightarrow Cu(NO_{3})_{2} + 2 Ag$

(D) In fuel cells,$Pt$ stands for platinum and $Pd$ stands for palladium. These metals are used as catalysts to increase the rate of the electrochemical reactions occurring at the electrodes.

(D) In a galvanic cell,oxidation occurs at the anode,where electrons are lost. Reduction occurs at the cathode,where electrons are gained. Therefore,the statement that the electrode at which electrons are lost is called the cathode is incorrect; it is actually called the anode.

(C) In secondary cells,electrode reactions can be reversed by an external electric energy source.
Therefore,these cells can be recharged by passing electric current and used again and again.
The current is passed in the direction opposite to that of the flow of current generated by the cell during discharge.

(B) During the charging process,the lead storage battery acts as an electrolytic cell.
At the anode (positive terminal during discharge,now connected to positive terminal of external source),$PbSO_{4}$ is oxidized to $PbO_{2}$:
$PbSO_{4(s)} + 2H_{2}O_{(l)} \rightarrow PbO_{2(s)} + S{O_{4}}^{2-}_{(aq)} + 4H^{+}_{(aq)} + 2e^{-}$
At the cathode (negative terminal during discharge,now connected to negative terminal of external source),$PbSO_{4}$ is reduced to $Pb$:
$PbSO_{4(s)} + 2e^{-} \rightarrow Pb_{(s)} + S{O_{4}}^{2-}_{(aq)}$
Therefore,$PbSO_{4}$ on the cathode is reduced to $Pb$.

(B) In a hydrogen-oxygen fuel cell,the reduction of oxygen occurs at the cathode.
The reaction is:
$O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^{-}_{(aq)}$

(A) The cell reaction is $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)}$.
Here, the number of electrons involved $(n)$ is $2$.
Using the relation $\Delta G^{\circ} = -n F E^{\circ}_{\text{cell}}$:
$-240,000 \text{ J} = -2 \times 96,500 \text{ C} \times E^{\circ}_{\text{cell}}$.
$E^{\circ}_{\text{cell}} = \frac{240,000}{2 \times 96,500} \approx 1.24 \text{ V}$.

(C) In a lead storage battery,during the discharge process,the anode is made of lead $(Pb)$.
At the anode,oxidation occurs where lead loses electrons to form lead sulfate $(PbSO_4)$.
The half-reaction at the anode is: $Pb_{\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} \rightarrow PbSO_{4\text{(s)}} + 2e^{-}$.
Therefore,option $C$ is the correct reaction occurring at the anode.

(A) The cell reaction is: $3 Ca_{(s)} + 2 Au^{3+}(aq) \rightarrow 3 Ca^{2+}(aq) + 2 Au_{(s)}$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Au^{3+}/Au} - E^{\circ}_{Ca^{2+}/Ca} = 1.50 \ V - (-2.87 \ V) = 4.37 \ V$
The number of electrons transferred $(n)$ is $6$.
Using the formula $\Delta G^{\circ} = -nFE^{\circ}_{cell}$:
$\Delta G^{\circ} = -6 \times 96500 \ C \ mol^{-1} \times 4.37 \ V$
$\Delta G^{\circ} = -2530230 \ J \ mol^{-1} = -2530.23 \ kJ \ mol^{-1}$
Rounding to three significant figures,$\Delta G^{\circ} = -2.53 \times 10^3 \ kJ \ mol^{-1}$.

(B) At the cathode,reduction of $Cu^{2+}$ ions occurs: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$; $E^{\circ}_{red} = 0.34 \ V$.
At the anode,oxidation of water occurs: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$; $E^{\circ}_{ox} = -1.23 \ V$.
The standard cell potential is $E^{\circ}_{cell} = E^{\circ}_{cathode} + E^{\circ}_{anode} = 0.34 \ V + (-1.23 \ V) = -0.89 \ V$.
Since the cell reaction is non-spontaneous $(E^{\circ}_{cell} < 0)$,the minimum external voltage required to drive the electrolysis is the magnitude of the cell potential,which is $0.89 \ V$.

(B) In a Daniell cell,the standard cell potential is $1.1 \ V$.
When an external voltage $E_{\text{external}}$ is applied such that $E_{\text{external}} > E_{\text{cell}}$,the cell acts as an electrolytic cell.
In this state,the direction of current flow reverses,moving from the cathode $(Cu)$ to the anode $(Zn)$.
Consequently,electrons flow from $Zn$ to $Cu$,and the chemical reaction becomes $Zn^{2+} + Cu \longrightarrow Zn + Cu^{2+}$.

(A) In the Daniell cell,$Zn$ acts as the anode and $Cu$ acts as the cathode.
At the anode,oxidation occurs: $Zn_{(s)} \longrightarrow Zn^{2+}_{(aq)} + 2e^-$.
At the cathode,reduction occurs: $Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$.
Since $Zn$ undergoes oxidation,it acts as a reducing agent.
Since $Cu^{2+}$ undergoes reduction,it acts as an oxidising agent.
The overall cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.
Comparing this with the given options,option $(A)$ states that $Zn$ is a reducing agent,which is correct. Note: Option $(D)$ in the original text had a typo in the reaction notation ($Zn^{+}Cu^{2+}$ instead of $Zn+Cu^{2+}$),but $(A)$ is fundamentally correct.

(B) The cell reaction is: $Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$
The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - (-0.76 \ V) = 1.56 \ V$
The number of electrons transferred in the reaction is $n = 2$.
Using the formula $\Delta G^{\circ} = -nFE^{\circ}$:
$\Delta G^{\circ} = -2 \times 96500 \ C/mol \times 1.56 \ V$
$\Delta G^{\circ} = -301080 \ J/mol = -301.08 \ kJ/mol \approx -301 \ kJ/mol$

(C) $KNO_3$ is used in a salt bridge because the ionic mobilities (velocities) of $K^{+}$ and $NO_3^{-}$ ions are nearly the same.
This ensures that the electrical neutrality of the two half-cells is maintained effectively without creating a liquid junction potential,as the ions do not accumulate at the electrodes.

(C) Statement $I$ is correct because a dry cell is a primary battery that cannot be recharged.
Statement $II$ is incorrect because in a dry cell,the zinc vessel acts as the anode (negative electrode),not the cathode.
Statement $III$ is correct because the electrolyte consists of a moist paste of $NH_4Cl$,$MnO_2$,and $ZnCl_2$ placed between the electrodes.
Statement $IV$ is correct because the standard potential of a dry cell is approximately $1.5 \ V$.
Therefore,the correct statements are $I, III, IV$.

(C) The mercury cell consists of a zinc-mercury amalgam as the anode and a paste of mercury$(II)$ oxide $(HgO)$ and carbon as the cathode.
The electrolyte used is a moist paste of potassium hydroxide $(KOH)$ and zinc oxide $(ZnO)$.

(A) In a Leclanché cell:
$1$. The anode is a zinc container $(Zn_{(s)} \rightarrow Zn^{2+} + 2e^-)$,so statement $(I)$ is correct.
$2$. The cathode is a graphite rod surrounded by powdered $MnO_2$ and carbon,so statement $(II)$ is correct.
$3$. The electrolyte is a moist paste of $NH_4Cl$ and $ZnCl_2$,not $ZnO$ and $KOH$. Thus,statement $(III)$ is incorrect.
$4$. The oxidation product involves $Zn^{2+}$ ions,not $ZnO$. Thus,statement $(IV)$ is incorrect.
Therefore,only statements $(I)$ and $(II)$ are correct.

(B) galvanic cell (also known as a voltaic cell) is an electrochemical device that converts the energy released by a spontaneous chemical reaction into electrical energy.

(B) The $Leclanche$ cell consists of a zinc container which acts as the anode.
The cathode is a carbon (graphite) rod surrounded by a mixture of powdered manganese dioxide $(MnO_2)$ and carbon.
The space between the cathode and anode is filled by a moist paste of ammonium chloride $(NH_4Cl)$ and zinc chloride $(ZnCl_2)$.

(A) According to Kohlrausch's law of independent migration of ions:
$(\Lambda_{m}^{\circ})_{CH_3-COOH} = (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{H^{+}}$
We have:
$(\Lambda_{m}^{\circ})_{(CH_3-COO)_2Ba} = 2(\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{Ba^{2+}} = x_1$
$(\Lambda_{m}^{\circ})_{BaCl_2} = (\Lambda_{m}^{\circ})_{Ba^{2+}} + 2(\Lambda_{m}^{\circ})_{Cl^{-}} = x_2$
$(\Lambda_{m}^{\circ})_{HCl} = (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}} = x_3$
To get $CH_3-COOH$,we perform the following operation:
$\frac{1}{2} [(\Lambda_{m}^{\circ})_{(CH_3-COO)_2Ba} - (\Lambda_{m}^{\circ})_{BaCl_2}] + (\Lambda_{m}^{\circ})_{HCl}$
$= \frac{1}{2} [2(\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{Ba^{2+}} - (\Lambda_{m}^{\circ})_{Ba^{2+}} - 2(\Lambda_{m}^{\circ})_{Cl^{-}}] + (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}}$
$= (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} - (\Lambda_{m}^{\circ})_{Cl^{-}} + (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}}$
$= (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{H^{+}} = (\Lambda_{m}^{\circ})_{CH_3-COOH}$
Therefore,the value is $\frac{x_1-x_2}{2} + x_3$.

(D) The reaction $Br_2 + 2Cl^- \longrightarrow 2Br^- + Cl_2$ does not take place because the standard cell potential $E^{\circ}_{cell}$ is negative,making the reaction non-spontaneous.
For this reaction:
Cathode: $Br_2 + 2e^- \longrightarrow 2Br^-$; $E^{\circ} = 1.09 \ V$
Anode: $2Cl^- \longrightarrow Cl_2 + 2e^-$; $E^{\circ} = 1.36 \ V$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.09 \ V - 1.36 \ V = -0.27 \ V$.
Since $E^{\circ}_{cell} < 0$,the reaction is non-spontaneous.
In contrast,$F_2$,$Cl_2$,and $Br_2$ can oxidize the halide ions below them in the group,resulting in positive $E^{\circ}_{cell}$ values for the other options.

(B) The cell reaction is:
Anode: $X \rightarrow X^{2+} + 2e^-$
Cathode: $2Y^+ + 2e^- \rightarrow 2Y$
Overall reaction: $X + 2Y^+ \rightarrow X^{2+} + 2Y$
Here,the number of electrons involved,$n = 2$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 \ V - (-1.7 \ V) = 2.5 \ V$.
The maximum work done is given by $W_{max} = -\Delta G^{\circ} = nFE^{\circ}_{cell}$.
$W_{max} = 2 \times 96500 \ C/mol \times 2.5 \ V = 482500 \ J/mol$.
Converting to $kJ/mol$: $W_{max} = \frac{482500}{1000} = 482.5 \ kJ/mol$.

(A) For a Daniell cell,the cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
Here,$n = 2$ (number of electrons transferred),
$F = 96500 \ C \ mol^{-1}$ (Faraday's constant),
$E^{\circ}_{cell} = 1.1 \ V$ (standard emf of the Daniell cell).
Substituting these values:
$\Delta G^{\circ} = -2 \times 96500 \ C \ mol^{-1} \times 1.1 \ V$
$\Delta G^{\circ} = -212300 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$:
$\Delta G^{\circ} = -212.3 \ kJ \ mol^{-1}$

(B) During the charging of a lead storage battery,the cell functions as an electrolytic cell.
The reactions occurring are the reverse of those during discharge.
At the anode,oxidation occurs where lead sulfate $(PbSO_4)$ is converted back into lead dioxide $(PbO_2)$.
The anode reaction is:
$PbSO_{4(s)} + 2H_2O_{(l)} \rightarrow PbO_{2(s)} + SO_{4(aq)}^{2-} + 4H_{(aq)}^+ + 2e^-$

(C) The extent of charge of a lead accumulator is determined by the specific gravity of the $H_2SO_4$ solution.
During discharge,$H_2SO_4$ is consumed,and its density (specific gravity) decreases.
During charging,$H_2SO_4$ is regenerated,and its density increases.
$A$ fully charged battery typically has a specific gravity of about $1.25$ to $1.30$.

(B) This question is based on Kohlrausch's law of independent migration of ions.
According to this law,at infinite dilution,the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions.
$\lambda^{\circ}_{MgSO_4} = \lambda^{\circ}_{Mg^{2+}} + \lambda^{\circ}_{SO_4^{2-}}$
Substituting the given values:
$\lambda^{\circ}_{MgSO_4} = 106 \ S \ cm^2 \ mole^{-1} + 160 \ S \ cm^2 \ mole^{-1} = 266 \ S \ cm^2 \ mole^{-1}$

(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
For $MgCl_2$,the dissociation is $MgCl_2 \rightarrow Mg^{2+} + 2Cl^{-}$.
Therefore,$\Lambda^{\circ}_{m(MgCl_2)} = \Lambda^{\circ}_{m(Mg^{2+})} + 2 \times \Lambda^{\circ}_{m(Cl^{-})}$.
Given: $\Lambda^{\circ}_{m(Mg^{2+})} = 106.0 \ S \ cm^2 \ mol^{-1}$ and $\Lambda^{\circ}_{m(Cl^{-})} = 76.3 \ S \ cm^2 \ mol^{-1}$.
$\Lambda^{\circ}_{m(MgCl_2)} = 106.0 + 2 \times 76.3 = 106.0 + 152.6 = 258.6 \ S \ cm^2 \ mol^{-1}$.

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(C)$ for strong electrolytes like $KCl$ is given by the Kohlrausch equation: $\Lambda_m = \Lambda_m^{\circ} - A \sqrt{C}$.
Here,$\Lambda_m^{\circ}$ is the molar conductivity at infinite dilution and $A$ is a constant.
This equation follows the linear form $y = mx + c$,where $y = \Lambda_m$,$x = \sqrt{C}$,$m = -A$ (slope),and $c = \Lambda_m^{\circ}$ (intercept).
Since the slope is negative $(-A)$,the plot of $\Lambda_m$ versus $\sqrt{C}$ is a straight line with a negative slope,indicating that $\Lambda_m$ decreases as $\sqrt{C}$ increases.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
For water $(H_2O)$,the dissociation is $H_2O \rightleftharpoons H^+ + OH^-$.
Therefore,$\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}$.
We can obtain this by combining strong electrolytes:
$\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
$\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
$\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$
By calculating $\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$,we get:
$(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-}) = \lambda^0_{H^+} + \lambda^0_{OH^-} = \Lambda^0_{m(H_2O)}$.
Thus,option $(B)$ is correct.

(B) Fuel cells (like the $H_2-O_2$ fuel cell) convert chemical energy directly into electrical energy.
They produce pure water as a byproduct,which is suitable for drinking,especially in space applications.
Therefore,the statement that water produced during the reaction cannot be used for drinking is incorrect.

(D) Electrical work done is given by $W = nFE_{cell}$,where $nF$ is the charge transferred and $E_{cell}$ is the potential difference.
Work done is the product of charge and potential difference $(W = q \times V)$,not the ratio.
Therefore,the statement in option $D$ is incorrect.

(A) reaction is spontaneous if $E^{0}_{cell} > 0$.
Given half-reactions are:
$Fe(OH)_{2}(s) + 2e^{-} \rightarrow Fe(s) + 2OH^{-}(aq)$ $(E^{0}_{red} = -0.88 \text{ V})$
$AgBr(s) + e^{-} \rightarrow Ag(s) + Br^{-}(aq)$ $(E^{0}_{red} = +0.07 \text{ V})$
For a spontaneous reaction,the half-reaction with the higher $E^{0}_{red}$ acts as the cathode (reduction) and the one with the lower $E^{0}_{red}$ acts as the anode (oxidation).
Anode (Oxidation): $Fe(s) + 2OH^{-}(aq) \rightarrow Fe(OH)_{2}(s) + 2e^{-}$ $(E^{0}_{ox} = +0.88 \text{ V})$
Cathode (Reduction): $2AgBr(s) + 2e^{-} \rightarrow 2Ag(s) + 2Br^{-}(aq)$ $(E^{0}_{red} = +0.07 \text{ V})$
Overall reaction: $Fe(s) + 2OH^{-}(aq) + 2AgBr(s) \rightarrow Fe(OH)_{2}(s) + 2Ag(s) + 2Br^{-}(aq)$
$E^{0}_{cell} = E^{0}_{cathode} + E^{0}_{ox} = 0.07 \text{ V} + 0.88 \text{ V} = 0.95 \text{ V}$.
Since $E^{0}_{cell} > 0$,the reaction is spontaneous. Thus,option $A$ is correct.