Electrochemical cells Questions in English

(B) In an electrochemical cell, the electrode where reduction takes place is defined as the cathode. Electrons flow towards the cathode, and it is typically the positive electrode in a galvanic cell. Therefore, reduction occurs at the cathode.

(A) reaction is spontaneous if the standard cell potential $E^0_{cell} > 0$. The formula is $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For $(A)$: $Ca$ is oxidized (anode), $Cd^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.403) - (-2.866) = +2.463 \text{ V}$. Since $E^0_{cell} > 0$, the reaction is spontaneous.
For $(B)$: $Br^-$ is oxidized (anode), $Sn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.136) - (1.08) = -1.216 \text{ V}$. Non-spontaneous.
For $(C)$: $Ag$ is oxidized (anode), $Ni^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.257) - (0.799) = -1.056 \text{ V}$. Non-spontaneous.
For $(D)$: $Au$ is oxidized (anode), $Zn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.763) - (1.68) = -2.443 \text{ V}$. Non-spontaneous.
Therefore, option $(A)$ is correct.

(C) $1$. Anode is the electrode where oxidation occurs (loss of electrons).
$2$. In option $(A)$, $F_2$ gains electrons (reduction).
$3$. In option $(B)$, $H^+$ and $O_2$ gain electrons (reduction).
$4$. In option $(C)$, $Fe^{2+}$ loses an electron to form $Fe^{3+}$, which is an oxidation process.
$5$. In option $(D)$, $Cu^{2+}$ gains electrons (reduction).
$6$. Therefore, only reaction $(C)$ represents oxidation and can occur at the anode.

(B) $1$. The standard reduction potential of the standard hydrogen electrode $(SHE)$ is $E^0_{(H^+|H_2)} = 0.00 \text{ V}$.
$2$. The standard reduction potential of the cadmium electrode is $E^0_{(Cd^{2+}|Cd)} = -0.403 \text{ V}$.
$3$. Since $E^0_{(H^+|H_2)} > E^0_{(Cd^{2+}|Cd)}$, the hydrogen electrode acts as the cathode (reduction) and the cadmium electrode acts as the anode (oxidation).
$4$. Anode reaction: $Cd(s) \rightarrow Cd^{2+}_{(aq)} + 2e^-$.
$5$. Cathode reaction: $2H^+_{(aq)} + 2e^- \rightarrow H_2(g)$.
$6$. Adding these two half-reactions gives the net cell reaction: $Cd(s) + 2H^+_{(aq)} \rightarrow Cd^{2+}_{(aq)} + H_2(g)$.

(A) Step $1$: $A$ voltaic cell (or galvanic cell) is an electrochemical cell that derives electrical energy from spontaneous redox reactions occurring within the cell.
Step $2$: In a voltaic cell, chemical energy is converted into electrical energy.
Step $3$: By convention, the anode is the electrode where oxidation occurs (negative terminal) and the cathode is the electrode where reduction occurs (positive terminal).
Step $4$: Therefore, option $A$ is the correct statement.

(C) Step $1$: $A$ mercury battery is a primary cell that uses a $Zn$ anode and a $HgO$ cathode.
Step $2$: The electrolyte used is a paste of $KOH$ and $ZnO$, which is strongly alkaline (basic), not acidic.
Step $3$: The cell reaction at the cathode is $HgO(s) + H_2O(l) + 2e^- \rightarrow Hg(l) + 2OH^-(aq)$, where $HgO$ is reduced to $Hg$.
Step $4$: Since the electrolyte is basic, the statement that it is strongly acidic is false.

(C) During the discharge of a lead storage battery, the following chemical reactions occur:
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Overall reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
From the overall reaction, it is clear that $H_2SO_4$ is consumed as the battery discharges.

(B) In a $H_2 - O_2$ fuel cell, the electrodes are made of porous carbon rods impregnated with a catalyst (like $Pt$ or $Pd$).
These electrodes are dipped into a hot aqueous solution of $KOH$ or $NaOH$, which acts as the electrolyte.
Therefore, the correct option is $KOH$.

(D) In a lead accumulator acting as a galvanic cell, the positive electrode is the cathode.
The reduction reaction at the cathode is: $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O$.
In $PbO_2$, the oxidation state of $Pb$ is $+4$.
In $PbSO_4$, the oxidation state of $Pb$ is $+2$.
Change in oxidation number = $+2 - (+4) = -2$.
Thus, the oxidation number decreases by $2$.

(D) During the recharging of a lead storage battery, the cell acts as an electrolytic cell.
At the cathode (negative electrode during recharging), the reduction reaction is: $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.
At the anode (positive electrode during recharging), the oxidation reaction is: $PbSO_4(s) + 2H_2O(\ell) \rightarrow PbO_2(s) + SO_4^{2-}(aq.) + 4H^+(aq.) + 2e^-$.
Therefore, the reaction occurring at the cathode is $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.

(C) $1$. $A$ salt bridge is a $U$-shaped tube containing an electrolyte (like $KCl$ or $KNO_3$) in a gel.
$2$. Its primary functions are to connect the two half-cells internally and to maintain electrical neutrality in the solutions of both half-cells by allowing the migration of ions.
$3$. It prevents the accumulation of charge, which would otherwise stop the flow of electrons and the reaction.
$4$. Therefore, option $C$ is the correct function.

(B) In an electrochemical cell, the cathode is the electrode where reduction (gain of electrons) takes place.
In a galvanic cell, the cathode is positive, while in an electrolytic cell, the cathode is negative.
Therefore, the statement 'Reduction occurs at the cathode' is always correct.

(C) $1$. The anode is the electrode where oxidation occurs.
$2$. Oxidation is defined as the loss of electrons.
$3$. In option $(A)$, $F_2$ gains electrons (reduction).
$4$. In option $(B)$, $H^+$ and $O_2$ gain electrons (reduction).
$5$. In option $(C)$, $Fe^{2+}$ loses an electron to form $Fe^{3+}$, which is an oxidation process.
$6$. In option $(D)$, $Cu^{2+}$ gains electrons (reduction).
$7$. Therefore, the reaction in option $(C)$ is the only one that represents oxidation and can occur at the anode.

(B) Step $1$: Identify the standard electrode potentials. For the standard hydrogen electrode $(SHE)$, $E^0(H^+|H_2) = 0.00 \text{ V}$. For the cadmium electrode, $E^0(Cd^{2+}|Cd) = -0.403 \text{ V}$.
Step $2$: Determine the anode and cathode. Since $E^0(Cd^{2+}|Cd) < E^0(H^+|H_2)$, the cadmium electrode acts as the anode (oxidation) and the hydrogen electrode acts as the cathode (reduction).
Step $3$: Write the half-reactions. Anode: $Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$. Cathode: $2H^+(aq) + 2e^- \rightarrow H_2(g)$.
Step $4$: Combine the half-reactions to get the net cell reaction: $Cd(s) + 2H^+(aq) \rightarrow Cd^{2+}(aq) + H_2(g)$.

(C) Step $1$: Henry's law states that the partial pressure of a gas in vapor phase is proportional to the mole fraction of the gas in the solution, given by $p = K_H x$. Thus, $(a) \rightarrow (ii)$.
Step $2$: Raoult's law states that for a solution of volatile liquids, the partial vapor pressure of each component is $p_1 = x_1 p_1^o$. Thus, $(b) \rightarrow (i)$.
Step $3$: The first law of thermodynamics states that the change in internal energy is the sum of heat and work, given by $\Delta U = q + w$. Thus, $(c) \rightarrow (iv)$.
Step $4$: Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its ions, given by $\Lambda_m^o = \nu_+ \lambda_+^o + \nu_- \lambda_-^o$. Thus, $(d) \rightarrow (iii)$.
Conclusion: The correct matching is $a – ii, b – i, c – iv, d – iii$.