Q.Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Concept and Intuition

This is a Daniell cell — the classic Zn–Cu galvanic cell. Normally, without any external source, electrons flow spontaneously from the Zn electrode (anode, oxidation) to the Cu electrode (cathode, reduction). The standard cell potential is about 1.1 V.

Here, an external voltage source $E_{\text{ext}}$ is inserted in the external circuit, but it is less than 1.1 V. That means the external source is trying to oppose the cell’s natural electron flow, but it is too weak to reverse it. The cell still operates as a galvanic cell (not an electrolytic cell), because the net driving force remains from Zn to Cu.

The Nernst equation helps us understand that even if concentrations change slightly, the cell potential remains positive as long as the reaction quotient $Q$ is less than the equilibrium constant $K$. Since $E_{\text{ext}} < E_{\text{cell}}$, the cell continues to discharge.

Step-by-step reasoning

1. Identify the spontaneous direction In a Zn–Cu cell, the standard reduction potentials are:

The cell reaction is:

with $E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}$.

Electrons flow from Zn (anode) to Cu (cathode) in the external circuit.

2. Effect of the external source $E_{\text{ext}} < 1.1\ \text{V}$

   The external source is connected opposing the cell’s natural polarity. However, because its voltage is less than the cell’s emf, the net potential difference still drives electrons from Zn to Cu.

   Tip

   Think of it like two batteries in series opposing: the larger one (the cell) wins, so current flows in its direction. Here the cell is the “larger” battery.

3. Evaluate Statement I: “Electrons flow from Cu electrode to Zn electrode.”

   The statement claims the opposite of the spontaneous direction. But as argued, the external source is too weak to reverse the flow. Therefore electrons still flow from Zn to Cu, not from Cu to Zn.

   Statement I is false as written — wait, careful: The statement says “from Cu electrode to Zn electrode”. That is the reverse of the actual direction. So Statement I is incorrect.

   Watch out

   A common mistake is to assume that any external voltage reverses the cell. But reversal only happens if $E_{\text{ext}} > E_{\text{cell}}$. Here $E_{\text{ext}} < 1.1$, so the cell still discharges normally.

4. Evaluate Statement II: “With increase in time, the weight of Zn electrode decreases and weight of Cu electrode increases.”

   In the spontaneous reaction:

   • Zn anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ (Zn dissolves, so its mass decreases).
   • Cu cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (Cu deposits, so its mass increases). This happens regardless of the external source as long as the cell is discharging. Since $E_{\text{ext}} < E_{\text{cell}}$, the cell continues to discharge, so Zn loses mass and Cu gains mass. Statement II is correct.

5. Combine the two statements

   • Statement I: Incorrect (electrons flow Zn → Cu, not Cu → Zn).
   • Statement II: Correct. Therefore the correct choice is (D).

Concept and Intuition: The Cell Representation and the Nernst Equation

This question tests your understanding of the direction of current and the mass changes at the electrodes in a working galvanic cell. The key is to remember that in a galvanic (voltaic) cell, the spontaneous redox reaction drives electrons through the external circuit. By convention, current flows opposite to electron flow (from positive to negative). The electrode where oxidation occurs (anode) loses mass; the electrode where reduction occurs (cathode) gains mass. The external opposing voltage $E_{\text{ext}} < 1.1\ \text{V}$ tells us the cell is still operating spontaneously (the cell’s emf is about 1.1 V for Zn–Cu under standard conditions).

Step-by-step reasoning

1. Identify the spontaneous reaction In a Zn–Cu cell, zinc is more reactive (higher reduction potential for Zn²⁺/Zn is –0.76 V, for Cu²⁺/Cu is +0.34 V). The spontaneous reaction is:

Zinc is oxidized (loses electrons), copper ions are reduced (gain electrons).

2. Determine electron flow and conventional current

   • Oxidation at the Zn electrode (anode): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ Electrons leave the Zn electrode and travel through the external circuit toward the Cu electrode.
   • Reduction at the Cu electrode (cathode): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ Electrons arrive at the Cu electrode.
   • Conventional current is defined as the flow of positive charge, opposite to electron flow. So electrons flow from Zn → Cu, meaning conventional current flows from Cu → Zn.
   • Statement I says: “Current flows from Cu electrode to Zn electrode.” This matches the conventional current direction. Statement I is correct.

3. Analyze mass changes at the electrodes

   • At the Zn anode: Zn metal is converted to Zn²⁺ ions, which go into solution. The solid Zn electrode loses mass.
   • At the Cu cathode: Cu²⁺ ions from solution gain electrons and deposit as solid Cu on the electrode. The Cu electrode gains mass.
   • Statement II says: “With increase in time the mass of Zn electrode increases and mass of Cu electrode decreases.” This is the exact opposite of what happens. Statement II is incorrect.

4. Check the role of the external voltage source

   The diagram shows $E_{\text{ext}} < 1.1\ \text{V}$. This is an opposing voltage, but because it is less than the cell’s emf, the cell still operates spontaneously (just with a reduced net voltage). The direction of electron flow and the mass changes remain the same as in an unopposed cell. So the presence of this small opposing voltage does not alter the correctness of the statements.

Concept and Intuition

The standard free energy change of a redox reaction is related to the cell's standard EMF by $\Delta G^\circ = -nFE^\circ_{cell}$, where $n$ is the total number of electrons transferred as balanced by the overall equation, and $F$ is Faraday's constant. A positive $E^\circ_{cell}$ (spontaneous reaction) corresponds to a negative $\Delta G^\circ$.

Step-by-Step Solution

1. Identify the half-reactions: reduction at cathode, $Au^{3+} + 3e^- \rightarrow Au$, $E^\circ = +1.50$ V; oxidation at anode, $Ca \rightarrow Ca^{2+} + 2e^-$, $E^\circ_{Ca^{2+}/Ca} = -2.87$ V.
2. Balance electrons: multiply the Au half-reaction by 2 and the Ca half-reaction by 3, giving $n = 6$ electrons transferred overall, matching the given equation $3Ca + 2Au^{3+} \rightarrow 3Ca^{2+} + 2Au$.
3. Compute $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.50 - (-2.87) = 4.37$ V.
4. Compute $\Delta G^\circ = -nFE^\circ_{cell} = -(6)(96500\ \text{C/mol})(4.37\ \text{V})$.
5. Calculate: $6 \times 96500 = 579000$; $579000 \times 4.37 = 2{,}530{,}230$ J $= 2530.23$ kJ. So $\Delta G^\circ = -2530.23$ kJ $\approx -2.53 \times 10^3$ kJ.

Common Mistakes

• Using $n = 3$ or $n = 2$ (from the individual half-reactions) instead of the correctly balanced $n = 6$ for the overall reaction.
• Sign errors — forgetting the negative sign in $\Delta G^\circ = -nFE^\circ$, or mishandling the negative anode potential when computing $E^\circ_{cell}$.
• Misplacing the decimal/power of ten when converting J to kJ (giving $10^4$ instead of $10^3$).

ANSWER: A

Concept and Intuition

The standard Gibbs energy change of a cell reaction is directly proportional to the number of moles of electrons transferred in the balanced overall reaction and the cell potential. Since the anode gives 3 electrons per P atom and the cathode needs 2 electrons per Q$^{2+}$ ion, the reaction must be balanced so both half-reactions transfer the same total number of electrons — found via the LCM of 3 and 2, which is 6.

Step-by-Step Solution

1. Anode (oxidation): $P \to P^{3+} + 3e^-$; Cathode (reduction): $Q^{2+} + 2e^- \to Q$.
2. Balance electrons: multiply the anode reaction by 2 and the cathode reaction by 3, so $n = 6$ electrons flow in the overall balanced cell reaction ($2P + 3Q^{2+} \to 2P^{3+} + 3Q$).
3. Apply $\Delta G^\circ = -nFE^\circ_{cell} = -6\times96500\,\text{C/mol}\times0.2\,\text{V}$.
4. Compute: $-6\times96500\times0.2 = -115800\,\text{J} = -115.8\,\text{kJ}$.

Common Mistakes

• Using $n=1$, $2$, or $3$ instead of the correctly balanced $n=6$.
• Forgetting to convert Joules to kilojoules.

ANSWER: A

Concept and Intuition

In a galvanic cell, a spontaneous ($\Delta G < 0$) redox reaction is split into separate oxidation (anode) and reduction (cathode) half-reactions connected by an external circuit, so the electron transfer that would otherwise happen directly (releasing energy as heat) instead flows through a wire and does electrical work. This is the reverse energy conversion of an electrolytic cell, which consumes electrical energy to drive a non-spontaneous chemical reaction.

Step-by-Step Solution

1. Identify the device: a galvanic/voltaic cell (e.g. Daniell cell) generates electricity from a spontaneous redox reaction.
2. The reaction proceeds because it is thermodynamically favourable (releases chemical energy).
3. That released energy is captured as an electric current through the external circuit rather than dissipated as heat.
4. Hence: chemical energy $\to$ electrical energy.

Common Mistakes

• Confusing a galvanic cell (chemical $\to$ electrical) with an electrolytic cell (electrical $\to$ chemical), which is the reverse process.

ANSWER: B

Concept and Intuition

In the cell notation $Cr/Cr^{3+}\,//\,Cd^{2+}/Cd$, the left electrode (written first) is the anode (oxidation), and the right electrode is the cathode (reduction). The cell potential is always cathode minus anode: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$. Once $E^\circ_{cell}$ is known, $\Delta G^\circ$ follows from the fundamental relation between electrical work and free energy, $\Delta G^\circ = -nFE^\circ_{cell}$, where $n$ is the total number of electrons transferred in the balanced overall cell reaction (found by matching the electron counts of the two half-reactions via their LCM).

Step-by-Step Solution

1. Identify electrodes: anode = $Cr^{3+}/Cr$ ($E^\circ=-0.74\ V$, oxidation: $Cr\to Cr^{3+}+3e^-$), cathode = $Cd^{2+}/Cd$ ($E^\circ=-0.40\ V$, reduction: $Cd^{2+}+2e^-\to Cd$).
2. Compute cell potential: $E^\circ_{cell} = E^\circ_{cathode}-E^\circ_{anode} = (-0.40)-(-0.74) = 0.34\ V$ (positive, confirming the reaction is spontaneous as written -- consistent with it being a genuine galvanic cell).
3. Balance the overall reaction by matching electrons: the anode releases 3 electrons per $Cr$, the cathode consumes 2 electrons per $Cd^{2+}$; the LCM of 3 and 2 is 6, so multiply the anode reaction by 2 and the cathode reaction by 3: $2Cr \to 2Cr^{3+}+6e^-$ and $3Cd^{2+}+6e^- \to 3Cd$, giving overall $2Cr+3Cd^{2+}\to 2Cr^{3+}+3Cd$ with $n=6$.
4. Apply $\Delta G^\circ = -nFE^\circ_{cell} = -6\times96{,}500\times0.34$.
5. Compute: $6\times96{,}500=579{,}000$; $579{,}000\times0.34=196{,}860\ J/mol \approx -196.8\ kJ/mol$ (negative, consistent with a spontaneous cell reaction).

Common Mistakes

• Reversing anode and cathode (using $E^\circ_{anode}-E^\circ_{cathode}$), which flips the sign of $E^\circ_{cell}$ and hence of $\Delta G^\circ$.
• Using $n=1$ or $n=2$ (from an unbalanced half-reaction) instead of the correctly balanced $n=6$ for the overall reaction, which would give a $\Delta G^\circ$ several times too small in magnitude.

ANSWER: C

Concept and Intuition

Electrochemical cell notation has a fixed convention: the electrode written on the left is the anode, where oxidation occurs, and the electrode on the right is the cathode, where reduction occurs — the double vertical line "||" represents the salt bridge separating the two half-cells. A positive emf value for the cell as written confirms the reaction is indeed spontaneous in the direction implied by this left-to-right convention (anode oxidation feeding electrons to cathode reduction).

Step-by-Step Solution

1. Left half-cell, $X\,|\,X^+(x\,M)$: this is the anode — oxidation occurs here, $X \to X^+ + e^-$.
2. Right half-cell, $Y^+(y\,M)\,|\,Y$: this is the cathode — reduction occurs here, $Y^+ + e^- \to Y$.
3. Combine the two half-reactions (balancing the single electron transferred in each): $X + Y^+ \to X^+ + Y$.
4. The given emf is $+0.5\,V$ (positive), which confirms this reaction as written is indeed the spontaneous direction (a negative emf would instead indicate the reverse reaction is spontaneous).

Common Mistakes

• Reversing which side is the anode/cathode — cell notation is standardized with anode always on the left, cathode on the right.
• Writing the overall reaction with charges/species that don't correspond to the actual redox half-reactions implied by the cell diagram (e.g. inventing an $X^-$ or $Y^-$ species, as in option D, which isn't consistent with the $X^+$/$Y^+$ notation given).

ANSWER: C