Electrochemical cells Questions in English

(D) The internal resistance of an electrochemical cell is determined by the nature of the electrolyte,the concentration of the electrolyte,the temperature,and the distance between the electrodes. It is independent of the $e.m.f.$ of the cell. Therefore,the correct option is $(D)$.

(D) process is considered non-spontaneous if it does not occur on its own and requires external work to proceed in a specific direction.
$A$. Dissolution of $CuSO_4$ in water is spontaneous.
$B$. The reaction between $H_2$ and $O_2$ to form water is spontaneous.
$C$. Water flowing down a hill is a spontaneous process due to gravity.
$D$. The flow of electric current from low potential to high potential is non-spontaneous because it requires an external power source (like a battery) to overcome the potential difference.

(C) Metals that are placed below hydrogen in the electrochemical series cannot displace hydrogen from dilute acids.
$Hg$ (Mercury) is placed below hydrogen in the electrochemical series,whereas $Ba$,$Pb$,and $Sn$ are placed above hydrogen.
Therefore,$Hg$ will not displace hydrogen.

(B) $Pb$ metal is used in the secondary cell,which is a lead storage battery.
Overall reaction in the battery: $Pb_{(s)} + PbO_{2(s)} + 2H_{2}SO_{4(aq)} \rightarrow 2PbSO_{4(s)} + 2H_{2}O_{(l)}$

(C) Lead storage cells (or lead-acid batteries) use an aqueous solution of $H_2SO_4$ (sulphuric acid) as the electrolyte.
During the discharge process,the lead anode is oxidized to $PbSO_4$ and the lead dioxide cathode is reduced to $PbSO_4$,consuming the sulphuric acid in the process.

(A) Copper metal $(Cu)$ can displace metals that are less reactive than it from their salt solutions.
According to the reactivity series,$Cu$ is more reactive than $Ag$ but less reactive than $Zn$,$Ba$,and $Na$.
When $Cu$ is added to $AgNO_3$ solution,the following displacement reaction occurs:
$Cu(s) + 2AgNO_3(aq) \to Cu(NO_3)_2(aq) + 2Ag(s)$
The formation of $Cu(NO_3)_2$ in the solution imparts a blue color to it.
Therefore,the correct option is $A$.

(D) In an electrolytic cell,an external power source (like a battery) is used to drive a non-spontaneous reaction.
Electrons are pumped by the external source into the cathode,where reduction occurs.
Electrons are removed from the anode,where oxidation occurs.
Therefore,the flow of electrons in the external circuit is from the anode to the cathode.

(B) The reaction for the electrolysis of $Al_2O_3$ is the reverse of the formation reaction: $\frac{2}{3} Al_2O_3 \to \frac{4}{3} Al + O_2$.
For this reaction,$\Delta G = +827 \ kJ \ mol^{-1} = 827000 \ J \ mol^{-1}$.
The number of electrons involved in the reaction is calculated from the oxidation state change: $Al^{3+} + 3e^- \to Al$. Since there are $\frac{4}{3}$ moles of $Al$,the total electrons $n = \frac{4}{3} \times 3 = 4$.
Using the formula $\Delta G = -nFE$,the minimum emf $E$ required is $E = -\frac{\Delta G}{nF}$.
$E = -\frac{827000 \ J \ mol^{-1}}{4 \times 96500 \ C \ mol^{-1}} = -2.14 \ V$.
Since we are looking for the minimum external potential to drive the non-spontaneous reaction,the magnitude is $2.14 \ V$.

(C) In a galvanic cell,the electrons flow from the anode to the cathode through the external circuit.
At the anode ($-ve$ pole),oxidation occurs,and at the cathode ($+ve$ pole),reduction occurs.

(D) The correct answer is $(D)$.
In a hydrogen-oxygen fuel cell,the theoretical efficiency is about $95 \%$,but practically,an efficiency of $60-70 \%$ has been attained.
Therefore,the statement that the efficiency is $23 \%$ is false.

(B) In a common dry cell (Leclanché cell):
Anode reaction: $Zn(s) \to Zn^{2+}(aq) + 2e^-$
Cathode reaction: $2MnO_2(s) + Zn^{2+}(aq) + 2e^- \to ZnMn_2O_4(s)$
Therefore,the reaction occurring at the cathode is $2MnO_2 + Zn^{2+} + 2e^- \to ZnMn_2O_4$.

(A) In a $Cu-Zn$ electrochemical cell (Daniell cell),the $Zn$ electrode acts as the anode and the $Cu$ electrode acts as the cathode.
At the anode,oxidation occurs: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$.
At the cathode,reduction occurs: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Since the standard reduction potential of $Cu^{2+}/Cu$ $(+0.34 \ V)$ is higher than that of $Zn^{2+}/Zn$ $(-0.76 \ V)$,$Cu$ acts as the cathode where reduction takes place.

(C) The reaction $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$ represents the overall cell reaction for a hydrogen-oxygen fuel cell.
In this cell,hydrogen and oxygen gases are supplied to produce electricity and water as the only product.

(C) The net cell reaction during discharge is: $Pb_{(s)} + PbO_{2_{(s)}} + 2H_2SO_{4_{(aq)}} \rightarrow 2PbSO_{4_{(s)}} + 2H_2O_{(l)}$.
During discharge,$H_2SO_4$ is consumed and its density decreases.
The net cell reaction during recharge is the reverse of the discharge reaction: $2PbSO_{4_{(s)}} + 2H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2_{(s)}} + 2H_2SO_{4_{(aq)}}$.
Thus,during the charging process,$H_2SO_4$ is regenerated,and its concentration increases.

(B) During the charging of a lead storage battery,the reactions at the electrodes are:
Anode: $PbSO_4 + 2e^- \to Pb + SO_4^{2-}$
Cathode: $PbSO_4 + 2H_2O \to PbO_2 + 4H^+ + SO_4^{2-} + 2e^-$
In both reactions,$H_2SO_4$ is regenerated,which increases the concentration of the electrolyte.

(C) In a dry cell (Leclanche cell),the reaction between $Zn$ and $NH_4Cl$ is as follows:
$Zn + 2NH_4Cl \to Zn(NH_3)_2Cl_2 + H_2 \uparrow$
Alternatively,the overall reaction involving the electrolyte can be represented as:
$Zn + 2NH_4Cl \to ZnCl_2 + 2NH_3 + H_2 \uparrow$
The colourless gas evolved is hydrogen $(H_2)$.

(A) In a galvanic cell,the anode (where oxidation occurs) is written on the $L.H.S.$ and the cathode (where reduction occurs) is written on the $R.H.S.$
For the given reaction: $Zn \to Zn^{2+} + 2e^-$ (Oxidation at anode) and $Cu^{2+} + 2e^- \to Cu$ (Reduction at cathode).
Thus,the cell representation is $Zn | Zn^{2+} || Cu^{2+} | Cu$.

(C) The correct answer is $C$.
In the electrochemical cell representation $Zn | Zn^{2+} || Cu^{2+} | Cu$,the left side represents the anode (negative electrode) and the right side represents the cathode (positive electrode).
Therefore,$Zn$ acts as the negative electrode where oxidation occurs $(Zn \rightarrow Zn^{2+} + 2e^-)$.

(A) galvanic cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. Therefore,it converts chemical energy into electrical energy. The correct option is $(A)$.

(B) In hydrogen-oxygen fuel cells,the reaction is $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$.
This reaction provides electrical energy (power) and the water produced is used as drinking water for astronauts.
Therefore,the correct answer is $B$.

(B) The cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.
Here,the number of electrons transferred,$n = 2$.
The standard cell potential,$E^0_{cell} = 1.10 \, V$.
The maximum work $(W_{max})$ obtained from a cell is equal to the decrease in Gibbs free energy $(-\Delta G^0)$.
Formula: $W_{max} = nFE^0_{cell}$.
Substituting the values: $W_{max} = 2 \times 96500 \, C/mol \times 1.10 \, V$.
$W_{max} = 212300 \, J/mol$.
Converting to $kJ$: $W_{max} = \frac{212300}{1000} \, kJ = 212.30 \, kJ$.

(B) In a Galvanic cell,the cathode is the electrode where reduction occurs.
For the cell $Zn|ZnSO_4||CuSO_4|Cu$,the cathode is the copper electrode.
The reduction reaction occurring at the copper cathode is: $Cu^{2+} + 2e^- \to Cu$.

(A) In a galvanic cell,the oxidation half-reaction occurs at the anode (left side) and the reduction half-reaction occurs at the cathode (right side).
The given reaction is: $Cu + 2Ag^{+} \to Cu^{+2} + 2Ag$.
Oxidation: $Cu_{(s)} \to Cu^{+2}_{(aq)} + 2e^-$.
Reduction: $2Ag^{+}_{(aq)} + 2e^- \to 2Ag_{(s)}$.
The cell notation is written as: $\text{Anode} | \text{Anode electrolyte} || \text{Cathode electrolyte} | \text{Cathode}$.
Therefore,the correct representation is $Cu_{(s)}|Cu^{+2}_{(aq)}||Ag^{+}_{(aq)}|Ag_{(s)}$.

(B) The cell representation $Zn_{(s)} | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu_{(s)}$ corresponds to the Daniel cell.
In this electrochemical cell,a $Zn$ rod is dipped in a $ZnSO_4$ solution and a $Cu$ rod is dipped in a $CuSO_4$ solution,connected by a salt bridge.

(C) The primary requirement for an electrolyte in a salt-bridge is that the transport numbers (or ionic mobilities) of the cation and anion should be approximately equal.
In a $KNO_3$ solution,the ionic velocities of $K^+$ and $NO_3^-$ are nearly the same.
This ensures that the junction potential is minimized and the electrical neutrality is maintained efficiently across the two half-cells.
Therefore,option $(C)$ is correct.

(B) In a standard electrochemical cell (Daniel cell type or similar),the electrode with a lower reduction potential acts as the anode,and the electrode with a higher reduction potential acts as the cathode.
For the $H_2/H^+$ and $Cu^{2+}/Cu$ system,the standard reduction potential of $Cu^{2+}/Cu$ is $+0.34 \ V$ and $H^+/H_2$ is $0.00 \ V$.
Since $E^{\circ}_{Cu^{2+}/Cu} > E^{\circ}_{H^+/H_2}$,$Cu$ acts as the cathode (where reduction occurs) and $H_2$ acts as the anode (where oxidation occurs).
Therefore,$H_2$ is the anode and $Cu$ is the cathode.

(C) In the given reaction,$Cu$ loses electrons to form $Cu^{2+}$,which is an oxidation process.
$Ag^{+}$ gains electrons to form $Ag$,which is a reduction process.
Therefore,the reduction half-cell reaction is $Ag^{+} + e^- \to Ag$.

(A) In a galvanic cell,the anode is the negative electrode where oxidation takes place,and the cathode is the positive electrode where reduction takes place.
Therefore,the statement '$A$' is incorrect because the anode is negative,not positive.

(B) In a lead storage battery,the anode is made of lead $(Pb)$.
During the discharge process,the lead at the anode undergoes oxidation.
The oxidation half-reaction is: $PbO_2(s)+4 H^{+}(aq)+SO_4{ }^{2-}(aq)+2 e^{-} \rightarrow PbSO_4(s)+2 H_2 O(l)$
Therefore,the correct option is $B$.

(B) In a dry cell (Leclanché cell),the zinc container acts as the anode.
At the anode,oxidation occurs,where zinc metal loses electrons to form zinc ions.
The reaction is: $Zn_{(s)} \to Zn^{2+} + 2e^-$.

(C) The anode is the electrode where the oxidation reaction occurs.
During oxidation,the species loses electrons.
Conversely,the cathode is the electrode where the reduction reaction occurs,which involves the gain of electrons.

(D) During the charging of a lead storage battery,the cell acts as an electrolytic cell.
At the cathode (negative electrode during charging),the reduction reaction is: $PbSO_4(s) + 2e^- \to Pb(s) + SO_4^{2-}(aq)$.
At the anode (positive electrode during charging),the oxidation reaction is: $PbSO_4(s) + 2H_2O(l) \to PbO_2(s) + SO_4^{2-}(aq) + 4H^{+}(aq) + 2e^-$.
Note: The provided option $(d)$ represents the reaction at the anode,not the cathode. However,based on standard multiple-choice conventions for this specific question,the intended answer is often the reaction occurring at the positive electrode (anode) during charging. If the question strictly asks for the cathode,the correct reaction is $PbSO_4 + 2e^- \to Pb + SO_4^{2-}$. Given the options,$(d)$ is the only reaction involving $PbO_2$ formation.

(B) The correct answer is $(B)$.
In a dry cell (Leclanché cell),$MnO_2$ (Manganese dioxide) acts as a depolarizer.
It is used to oxidize the hydrogen gas produced at the cathode to water,thereby preventing the polarization of the cell.

(D) The correct option is $(D)$.
During the discharge of a lead storage battery,the chemical reaction is:
$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
As shown in the equation,$H_2SO_4$ is consumed during the discharge process.

(D) The electrochemical series indicates the reactivity of metals. $A$ metal placed lower in the electrochemical series cannot displace a metal placed higher in the series from its salt solution.
Since $Cu$ is placed below $Al$ in the electrochemical series,$Cu$ is less electropositive than $Al$.
Therefore,$Cu$ cannot displace $Al$ from $Al(NO_3)_3$ solution.
Thus,there is no reaction.

(C) In an electrochemical cell,chemical energy is converted into electrical energy through spontaneous redox reactions.

(A) The primary functions of a salt bridge in a galvanic cell are to complete the electrical circuit by allowing the flow of ions between the two half-cells and to maintain electrical neutrality in the solutions. Therefore,the correct option is $(A)$.

(D) The correct answer is $D$.
According to the electrochemical series,the standard reduction potential of $Cu^{2+}/Cu$ is $E^o = +0.34 \ V$ and that of $Fe^{2+}/Fe$ is $E^o = -0.44 \ V$.
Since $Cu$ has a higher reduction potential than $Fe$,$Cu$ cannot displace $Fe$ from its salt solution.
Therefore,no reaction will take place when a $Cu$ strip is placed in a $FeSO_4$ solution.

(A) salt bridge is typically constructed using an inert electrolyte whose ions have similar transport numbers,such as $KCl$,$KNO_3$,or $NH_4NO_3$.
$CH_3COOK$ (potassium acetate) is not commonly used because the acetate ion $(CH_3COO^-)$ can undergo hydrolysis or react with other ions in the solution,potentially affecting the cell potential or causing precipitation.

(D) The standard reference electrode,known as the calomel electrode,is constructed using mercury $(Hg)$,mercurous chloride $(Hg_2Cl_2)$,and a potassium chloride $(KCl)$ solution.

(B) In a hydrogen-oxygen fuel cell,the following reactions take place to create a potential difference between the two electrodes:
Anode: $2H_{2(g)} + 4OH^{-}_{(aq)} \to 4H_2O_{(l)} + 4e^{-}$
Cathode: $O_{2(g)} + 2H_2O_{(l)} + 4e^{-} \to 4OH^{-}_{(aq)}$
Overall reaction: $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$
The net reaction is equivalent to the combustion of hydrogen to form water,which facilitates the flow of electrons and creates a potential difference.

(C) In a $Daniel$ cell,the copper electrode acts as the cathode. \\ During the operation of the cell,cations in the electrolyte move toward the cathode (copper electrode) to maintain electrical neutrality,where reduction takes place. \\ Therefore,statement $(c)$ is correct.

(B) The standard reduction potential of $Cu^{2+}/Cu$ is $+0.34 \ V$,while for $H^+/H_2$ it is $0.00 \ V$.
Since the reduction potential of $H_2$ is lower than that of $Cu$,$H_2$ acts as the anode (where oxidation occurs) and $Cu$ acts as the cathode (where reduction occurs).
Therefore,the correct statement is that $H_2$ is the anode and $Cu$ is the cathode.

(D) Fuel cells are electrochemical devices that convert the energy of combustion of fuels like $H_2$,$CH_4$,or $CH_3OH$ directly into electrical energy.
$1$. They are more efficient than conventional methods of producing electricity.
$2$. They are free from pollution as the byproduct is usually water.
$3$. They run continuously as long as the reactants are supplied to the electrodes.
Therefore,all the given statements are true.

(A) The correct option is $(A)$.
Dilute $H_2SO_4$ (sulfuric acid) is used as the electrolyte in a lead storage battery.

(C) The given cell reaction is $2AgCl_{(s)} + H_{2(g)} \to 2HCl_{(aq)} + 2Ag_{(s)}$.
In this reaction,$H_2$ is oxidized to $H^+$ at the anode,and $AgCl$ is reduced to $Ag$ at the cathode.
The anode is a standard hydrogen electrode represented as $Pt_{(s)} | H_{2(g)}, 1 \ bar | H^+_{(aq)}$.
The cathode involves the $Ag/AgCl$ electrode,represented as $Cl^-_{(aq)} | AgCl_{(s)} | Ag_{(s)}$.
Combining these,the cell notation is $Pt_{(s)} | H_{2(g)}, 1 \ bar | 1 \ M \ HCl_{(aq)} | AgCl_{(s)} | Ag_{(s)}$.

(C) During electrolysis,the cation with the highest standard reduction potential is reduced first at the cathode.
Comparing the given reduction potentials: $E^0_{Ag^{+}/Ag} (+0.80 \ V) > E^0_{Hg_2^{2+}/Hg} (+0.79 \ V) > E^0_{Cu^{2+}/Cu} (+0.34 \ V) > E^0_{Mg^{2+}/Mg} (-2.37 \ V)$.
Therefore,the order of deposition is $Ag$,then $Hg$,then $Cu$.
$Mg^{2+}$ ions are not reduced because the reduction of water $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$ occurs at a more positive potential than the reduction of $Mg^{2+}$ in an aqueous solution.
Thus,the sequence of deposition is $Ag, Hg, Cu$.

(A) The reaction occurring is: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
Since the standard reduction potential of $Ag^+/Ag$ $(+0.80 \ V)$ is greater than that of $Cu^{2+}/Cu$ $(+0.34 \ V)$,$Cu$ undergoes oxidation to form $Cu^{2+}$ ions.
The blue colour of the solution is characteristic of the presence of hydrated $Cu^{2+}$ ions in the aqueous medium.

(A) The chemical reactivity of metals with water is determined by their position in the electrochemical series. Metals with more negative standard reduction potentials are more reactive. The standard reduction potentials $(E^{\circ})$ are: $K^+/K = -2.93 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Zn^{2+}/Zn = -0.76 \ V$,and $Cu^{2+}/Cu = +0.34 \ V$. Since the reactivity decreases as the reduction potential increases,the correct order of reactivity is $K > Mg > Zn > Cu$.

(C) The Saturated Calomel Electrode $(SCE)$ is a common reference electrode used in electrochemistry.
It consists of mercury $(Hg)$ in contact with a paste of mercurous chloride $(Hg_2Cl_2)$ and a saturated solution of potassium chloride $(KCl)$.
Therefore,the substance used in the construction of this electrode is $Hg_2Cl_2$ (mercurous chloride).