Q.Depict the galvanic cell in which the reaction $Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$ takes place. Further show:

The key to this question is understanding the thermodynamic relationship between cell potential and Gibbs free energy. A spontaneous process in a galvanic cell means the cell can do electrical work on its surroundings without external input. The sign conventions are the critical link.

1. Recall the fundamental equation connecting Gibbs free energy and cell potential:

where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant, and $E_{\text{cell}}$ is the cell potential under the given conditions.

For a spontaneous reaction, $\Delta G < 0$. Since $n$ and $F$ are positive, the negative sign forces $E_{\text{cell}} > 0$ for spontaneity.

2. Now consider standard conditions (1 M concentrations, 1 atm pressure, 298 K). The equation becomes:

Spontaneity under standard conditions requires $\Delta G^\circ < 0$, which again implies $E^\circ_{\text{cell}} > 0$.

So the pair $(E^\circ_{\text{cell}} > 0, \Delta G^\circ < 0)$ is always true for a spontaneous cell reaction under standard conditions.

3. Examine each option:
   • (A) $E^\circ_{\text{cell}} > 0$; $\Delta G^\circ < 0$ — matches the reasoning above.
   • (B) $E^\circ_{\text{cell}} = 0$; $\Delta G^\circ = 0$ — this describes equilibrium, not spontaneity.
   • (C) $E^\circ_{\text{cell}} < 0$; $\Delta G^\circ > 0$ — this is non-spontaneous (electrolytic cell under standard conditions).
   • (D) $E^\circ_{\text{cell}} > 0$; $\Delta G^\circ > 0$ — impossible because $\Delta G^\circ$ and $E^\circ_{\text{cell}}$ always have opposite signs.

Step 1 — The half-cell and the Nernst equation.

The reduction half-reaction is

The Nernst equation for a reduction electrode at $298\,\mathrm{K}$ is

Step 2 — Substitute the species.

Aluminium metal is a pure solid, so its activity is $1$ and it does not appear in the quotient. The oxidised form is $\mathrm{Al^{3+}}$:

Using $\log(1/x) = -\log x$, this simplifies to

Step 3 — Read off the dependence.

$\log[\mathrm{Al^{3+}}]$ is a monotonically increasing function of $[\mathrm{Al^{3+}}]$, and it carries a positive coefficient $\tfrac{0.059}{3}$. Therefore as $[\mathrm{Al^{3+}}]$ increases, $E$ increases.

Step 4 — The physical reason (why this must be so).

By Le Chatelier's principle, adding more $\mathrm{Al^{3+}}$ pushes the reduction $\mathrm{Al^{3+}} + 3e^- \to \mathrm{Al}$ forward. A greater tendency to be reduced is a higher (more positive) reduction potential. The algebra and the chemistry agree.

Numerical check: raising $[\mathrm{Al^{3+}}]$ from $0.1\,\mathrm{M}$ to $1\,\mathrm{M}$ changes the log term from $-1$ to $0$, so $E$ rises by $\tfrac{0.059}{3}(0 - (-1)) \approx +0.0197\,\mathrm{V}$ — an increase, as claimed.

ANSWER: D

The key concept here is the relationship between the standard cell potential ($E^\circ_{\text{cell}}$) and the equilibrium constant ($K_c$) via the Nernst equation. At equilibrium, the cell potential is zero, and the reaction quotient $Q$ equals $K_c$. The equation is:

where $n$ is the number of electrons transferred, $F$ is Faraday’s constant, $R$ is the gas constant, and $T$ is temperature in Kelvin.

A common pitfall: students often forget that the equilibrium constant for the reverse reaction is simply the reciprocal of that for the forward reaction. Also, careful attention to the sign of $E^\circ$ and the value of $n$ is essential.

Let’s work through it step by step.

1. Identify $n$, the number of electrons transferred.

   In the forward reaction:

   $4 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Br}_2 + 4e^-$ (oxidation)

   $\mathrm{O}_2 + 4 \mathrm{H}^{+} + 4e^- \rightarrow 2 \mathrm{H}_2\mathrm{O}$ (reduction)

   So $n = 4$.

2. Write the Nernst equation at equilibrium for the forward reaction.

   At 298 K, using base-10 logarithms:

(This comes from $\frac{2.303 RT}{F} \approx 0.0591$ at 298 K.)

3. Plug in the given $E^\circ_{\text{cell}} = 0.16$ V and $n=4$.

4. Solve for $K_c$ of the forward reaction.

(A more precise calculation using $0.05916$ gives $K_c \approx 7.03 \times 10^{10}$, matching option D’s value for the forward reaction.)

5. Now consider the reverse reaction. The reverse reaction is: $2 \mathrm{Br}_2 + 2 \mathrm{H}_2\mathrm{O} \rightarrow 4 \mathrm{Br}^{-} + \mathrm{O}_2 + 4 \mathrm{H}^{+}$ Its equilibrium constant $K_c'$ is the reciprocal of the forward $K_c$:

6. Match with the options. The value $1.422 \times 10^{-11}$ corresponds exactly to option (C).

The key idea is the Nernst equation, which tells us how the cell potential depends on the concentrations of the ions involved. For a spontaneous cell, the EMF is largest when the reaction quotient $Q$ is smallest — that is, when the reactants are concentrated and the products are dilute. Here, Al³⁺ is a product and Fe²⁺ is a reactant, so we want to increase Fe²⁺ or decrease Al³⁺ to raise the EMF.

Let’s work through it step by step.

1. Write the cell reaction. The cell notation is:

The left half-cell is the anode (oxidation):

The right half-cell is the cathode (reduction):

To balance electrons, multiply the Al half-reaction by 2 and the Fe half-reaction by 3:

So the overall reaction has Al³⁺ as a product and Fe²⁺ as a reactant.

2. Apply the Nernst equation. For the reaction $aA + bB \rightarrow cC + dD$, the Nernst equation at 298 K is:

where $Q = \frac{[\text{products}]^c [\text{products}]^d}{[\text{reactants}]^a [\text{reactants}]^b}$.

Here, $n = 6$ (the total electrons transferred), and:

(Solids Al and Fe have activity = 1, so they don’t appear.)

3. See how E changes with concentration. The EMF is:

To increase $E$, we want the term $-\frac{0.0591}{6} \log Q$ to become less negative (or more positive). That means we want $\log Q$ to decrease — i.e., make $Q$ smaller.

• $Q$ gets smaller if $[\mathrm{Al}^{3+}]$ decreases (product concentration down).
• $Q$ gets smaller if $[\mathrm{Fe}^{2+}]$ increases (reactant concentration up).

4. Check the options.
   • (A) Increase both: $[\mathrm{Al}^{3+}] \uparrow$ makes $Q$ larger → EMF decreases.
   • (B) Increase only $[\mathrm{Al}^{3+}]$: same effect → EMF decreases.
   • (C) Increase only $[\mathrm{Fe}^{2+}]$: $Q$ smaller → EMF increases. ✓
   • (D) Decrease both: $[\mathrm{Fe}^{2+}] \downarrow$ makes $Q$ larger → EMF decreases.

The key concept here is the Nernst equation, which adjusts the standard cell potential ($E^\circ$) for non-standard conditions (concentrations and gas pressures). The reaction involves a solid cadmium electrode, hydrogen ions, and hydrogen gas, so we treat the cell as a concentration cell with a redox couple. The intuition: even though $E^\circ$ for the Cd²⁺/Cd half-cell is given, the overall cell reaction combines it with the standard hydrogen electrode (SHE) under non-standard conditions. The Nernst equation lets us compute the actual voltage.

1. Identify the half-reactions and standard cell potential.

   The overall reaction is:

   $\mathrm{Cd(s)} + 2\mathrm{H}^+ \rightarrow \mathrm{Cd}^{2+} + \mathrm{H}_2(g)$.

   The half-reactions are:

   • Oxidation: $\mathrm{Cd(s)} \rightarrow \mathrm{Cd}^{2+} + 2e^-$ with $E^\circ_{\text{ox}} = +0.4\ \mathrm{V}$ (since $E^\circ(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.4\ \mathrm{V}$ for reduction, oxidation reverses the sign).
   • Reduction: $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2(g)$ with $E^\circ_{\text{red}} = 0\ \mathrm{V}$ (standard hydrogen electrode). The standard cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = 0 + 0.4 = 0.4\ \mathrm{V}$.

2. Write the Nernst equation for the cell.

   For the reaction $aA + bB \rightarrow cC + dD$, the Nernst equation at 298 K is:

where $n$ is the number of electrons transferred (here $n=2$), and $Q$ is the reaction quotient.

For our reaction:

Note: Solids (Cd) and liquids (if any) have activity = 1, so they don't appear.

3. Plug in the given values. $[\mathrm{Cd}^{2+}] = 0.01\ \mathrm{M}$, $P_{\mathrm{H}_2} = 0.8\ \mathrm{atm}$, $[\mathrm{H}^+] = 0.02\ \mathrm{M}$. So:

4. Compute the logarithm.

   $\log_{10}(20) = \log_{10}(2 \times 10) = \log_{10}2 + 1 \approx 0.3010 + 1 = 1.3010$

5. Apply the Nernst equation.

First, $\frac{0.0591}{2} = 0.02955$.

Then $0.02955 \times 1.3010 \approx 0.03845$ (since $0.03 \times 1.301 = 0.03903$, but more precisely: $0.02955 \times 1.3010 = 0.03845$).

So $E = 0.4 - 0.03845 = 0.36155\ \mathrm{V}$, which rounds to $0.3616\ \mathrm{V}$.

1. The Nernst equation for this electrode

The half-reaction is a one-electron reduction:

(The solid Ag has unit activity, so it does not appear in the quotient.)

2. The qualitative rule that follows

Because $\log[\mathrm{Ag^{+}}]$ is an increasing function of $[\mathrm{Ag^{+}}]$:

The higher the concentration of the oxidised species ($\mathrm{Ag^{+}}$), the higher (more positive) the reduction potential.

This makes chemical sense — more $\mathrm{Ag^{+}}$ in solution drives the reduction to Ag forward (Le Chatelier).

3. Compute each electrode ($E^{\circ} = +0.80\ \mathrm{V}$)

4. The order

Electrodes A and D are identical (both 0.001 M), so their potentials are equal; any correct ordering must place them together at the bottom, below C, which is below B. Only option (A) — B > C > D > A — has that structure. Options (C) and (D) invert the trend entirely (putting the most dilute first), and (B) puts the most concentrated electrode last, both contradicting the Nernst equation.

ANSWER: A

Concept: Nernst equation for a galvanic cell.

Cell: Fe | Fe^2+ (0.001 M) || H^+ (0.01 M) | H2 (1 bar) | Pt

Anode (oxidation): Fe -> Fe^2+ + 2e^-

Cathode (reduction): 2 H^+ + 2e^- -> H2

Overall: Fe + 2 H^+ -> Fe^2+ + H2 , n = 2

E(cell,std) = E(cathode) - E(anode) = 0.00 - (-0.44) = +0.44 V

Reaction quotient:

Q = [Fe^2+] * p(H2) / [H^+]^2 = (0.001)(1) / (0.01)^2 = 0.001 / 0.0001 = 10

Nernst:

E = E(std) - (0.0591/n) log Q

= 0.44 - (0.0591/2) log 10

= 0.44 - 0.0296

= 0.4104 V ~ 0.41 V

ANSWER: C

Step 1 — Decide which half-cell is the cathode.

Given:

In a galvanic cell the electrode with the higher (more positive) reduction potential acts as the cathode (reduction), and the other is the anode (oxidation) — this is what makes $E^\circ_{cell}$ positive and the reaction spontaneous.

Since $0.76 > 0.55$:

• Cathode (reduction): $Fe^{3+} + e^- \longrightarrow Fe^{2+}$
• Anode (oxidation): $2I^- \longrightarrow I_2 + 2e^-$

Step 2 — Balance the electrons to get $n$.

The iodide half-reaction releases 2 electrons, so the iron half-reaction must be doubled:

Step 3 — Compute $E^\circ_{cell}$.

(Note: $E^\circ$ is an intensive property — it is not multiplied when the half-reaction is doubled. Only $n$ changes.)

Step 4 — Link $E^\circ_{cell}$ to the equilibrium constant.

At equilibrium the cell is dead ($E_{cell} = 0$, $Q = K_c$), and the Nernst equation gives the standard relation

using the value $\dfrac{2.303RT}{F} = 0.06$ supplied in the question.

Step 5 — Substitute.

Step 6 — Sanity check. $E^\circ_{cell} > 0 \Rightarrow \Delta G^\circ < 0 \Rightarrow K_c \gg 1$, i.e. the forward reaction is strongly favoured — consistent with $10^7$. ✓

ANSWER: B

1. The key principle — why you cannot just subtract the potentials. $E^\circ$ is an intensive quantity (energy per electron), so potentials of different half-reactions do not add. The extensive quantity that does add is the Gibbs free energy:

This is the single idea the whole question is testing.

2. Write the three half-reactions with their electron counts $n$.