Q.Write the Nernst equation and emf of the following cells at 298 K:

Concept: Cell Representation & Nernst Equation

The Nernst equation for a cell reaction $aA + bB \rightarrow cC + dD$ at 298 K is:

where $n$ is the number of electrons transferred, and concentrations are in molarity (gases in bar, pure solids/liquids = 1).

Step 1: Identify half-reactions, find $E^\circ$ from standard reduction potentials, and determine $n$.

Step 2: Write the cell reaction and the reaction quotient $Q$.

Step 3: Substitute into Nernst equation at 298 K.

(i) $Mg(s)|Mg^{2+}(0.001\ M)||Cu^{2+}(0.0001\ M)|Cu(s)$

Anode: $Mg \rightarrow Mg^{2+} + 2e^-$

Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$

$E^\circ_{\text{cell}} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Mg^{2+}/Mg} = 0.34 - (-2.37) = 2.71\ \text{V}$, $n=2$

$E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]} = 2.71 - 0.02955 \log \frac{0.001}{0.0001}$ …

The Core Idea

A cell’s emf depends not just on the metals involved but on how concentrated the ions are. The Nernst equation captures this:

At 298 K, using $\log_{10}$, this becomes:

Here $n$ is the number of electrons transferred in the balanced cell reaction, and $Q$ is the reaction quotient (products over reactants, solids and pure liquids omitted, gases in bar, ions in molarity).

The trick: always write the spontaneous cell reaction first. The left electrode is the anode (oxidation), the right is the cathode (reduction). Then $Q$ follows naturally.

(i) $Mg(s)|Mg^{2+}(0.001\ M)||Cu^{2+}(0.0001\ M)|Cu(s)$

1. Identify half‑reactions and $E^\circ$

Anode (oxidation): $Mg(s) \rightarrow Mg^{2+} + 2e^-$

Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu(s)$

Standard reduction potentials (from tables):

• $E^\circ_{Cu^{2+}/Cu} = +0.34\ \text{V}$
• $E^\circ_{Mg^{2+}/Mg} = -2.37\ \text{V}$

So $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-2.37) = 2.71\ \text{V}$.

2. Write the net cell reaction

$Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$

Electrons transferred: $n = 2$.

3. Reaction quotient $Q$

4. Apply Nernst equation

(ii) $Fe(s)|Fe^{2+}(0.001\ M)||H^+(1\ M)|H_2(g)(1\ bar)|Pt(s)$

1. Half‑reactions and $E^\circ$

Anode: $Fe(s) \rightarrow Fe^{2+} + 2e^-$

Cathode: $2H^+ + 2e^- \rightarrow H_2(g)$

$E^\circ_{Fe^{2+}/Fe} = -0.44\ \text{V}$, $E^\circ_{H^+/H_2} = 0.00\ \text{V}$ (by definition).

So $E^\circ_{\text{cell}} = 0.00 - (-0.44) = 0.44\ \text{V}$.

2. Net reaction

$Fe(s) + 2H^+(aq) \rightarrow Fe^{2+}(aq) + H_2(g)$

$n = 2$.

3. $Q$

4. Nernst

(iii) $Sn(s)|Sn^{2+}(0.050\ M)||H^+(0.020\ M)|H_2(g)(1\ bar)|Pt(s)$

1. Half‑reactions and $E^\circ$

Anode: $Sn(s) \rightarrow Sn^{2+} + 2e^-$

Cathode: $2H^+ + 2e^- \rightarrow H_2(g)$

$E^\circ_{Sn^{2+}/Sn} = -0.14\ \text{V}$, $E^\circ_{H^+/H_2} = 0.00\ \text{V}$.

$E^\circ_{\text{cell}} = 0.00 - (-0.14) = 0.14\ \text{V}$.

2. Net reaction

$Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$

$n = 2$.

3. $Q$

4. Nernst

$\log(125) = \log(5^3) = 3 \log 5 \approx 3 \times 0.6990 = 2.097$

(iv) $Pt(s)|Br^-(0.010\ M)|Br_2(l)||H^+(0.030\ M)|H_2(g)(1\ bar)|Pt(s)$

1. Half‑reactions and $E^\circ$

Anode (oxidation): $2Br^- \rightarrow Br_2(l) + 2e^-$

Cathode (reduction): $2H^+ + 2e^- \rightarrow H_2(g)$ …

Method: Nernst Equation for Cell EMF

This method uses the Nernst equation to calculate the cell potential under non-standard conditions. The standard approach is:

Steps:

1. Identify half-reactions — oxidation (anode, left) and reduction (cathode, right)
2. Write the overall cell reaction
3. Find $E^\circ_{\text{cell}}$ using standard reduction potentials
4. Apply the Nernst equation at 298 K:

where $n$ = number of electrons transferred, $Q$ = reaction quotient

(i) $Mg(s)|Mg^{2+}(0.001\ M)||Cu^{2+}(0.0001\ M)|Cu(s)$

Half-reactions:

• Anode (oxidation): $Mg(s) \rightarrow Mg^{2+} + 2e^-$
• Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu(s)$

Overall: $Mg(s) + Cu^{2+} \rightarrow Mg^{2+} + Cu(s)$, $n = 2$

Standard potentials:

• $E^\circ_{Mg^{2+}/Mg} = -2.37\ \text{V}$
• $E^\circ_{Cu^{2+}/Cu} = +0.34\ \text{V}$
• $E^\circ_{\text{cell}} = 0.34 - (-2.37) = +2.71\ \text{V}$

Nernst equation:

Answer: $\boxed{E_{\text{cell}} = 2.68\ \text{V}}$

(ii) $Fe(s)|Fe^{2+}(0.001\ M)||H^+(1\ M)|H_2(g)(1\ bar)|Pt(s)$

Half-reactions:

• Anode: $Fe(s) \rightarrow Fe^{2+} + 2e^-$
• Cathode: $2H^+ + 2e^- \rightarrow H_2(g)$

Overall: $Fe(s) + 2H^+ \rightarrow Fe^{2+} + H_2(g)$, $n = 2$

Standard potentials:

• $E^\circ_{Fe^{2+}/Fe} = -0.44\ \text{V}$
• $E^\circ_{H^+/H_2} = 0.00\ \text{V}$
• $E^\circ_{\text{cell}} = 0.00 - (-0.44) = +0.44\ \text{V}$

Nernst equation:

Answer: $\boxed{E_{\text{cell}} = 0.529\ \text{V}}$

(iii) $Sn(s)|Sn^{2+}(0.050\ M)||H^+(0.020\ M)|H_2(g)(1\ bar)|Pt(s)$

Half-reactions:

• Anode: $Sn(s) \rightarrow Sn^{2+} + 2e^-$
• Cathode: $2H^+ + 2e^- \rightarrow H_2(g)$

Overall: $Sn(s) + 2H^+ \rightarrow Sn^{2+} + H_2(g)$, $n = 2$

Standard potentials:

• $E^\circ_{Sn^{2+}/Sn} = -0.14\ \text{V}$
• $E^\circ_{H^+/H_2} = 0.00\ \text{V}$
• $E^\circ_{\text{cell}} = 0.00 - (-0.14) = +0.14\ \text{V}$

Nernst equation:

$\log 125 = \log(5^3) = 3\log 5 = 3(0.6990) = 2.097$ …

Common Mistakes in Cell Representation & Nernst Equation Problems

1. Writing the Cell Reaction in the Wrong Direction

The Mistake: Students often write the anode reaction as reduction or the cathode reaction as oxidation.

How to Avoid:

• Left = Anode (Oxidation) — Always remember: the left side of the cell notation is where oxidation occurs.
• Right = Cathode (Reduction) — The right side is where reduction occurs.
• Write the half-reactions separately first:
  • Anode (oxidation): $M \rightarrow M^{n+} + ne^-$
  • Cathode (reduction): $M^{n+} + ne^- \rightarrow M$

2. Forgetting to Balance Electrons in the Overall Reaction

The Mistake: Adding half-reactions without ensuring the number of electrons lost equals electrons gained.

How to Avoid:

• Count electrons in each half-reaction.
• Multiply half-reactions by appropriate integers so $n$ is the same.
• Example: For $Mg|Mg^{2+}||Cu^{2+}|Cu$:
  • Anode: $Mg \rightarrow Mg^{2+} + 2e^-$
  • Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$
  • Overall: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$ (electrons already balanced)

3. Using Wrong $n$ in the Nernst Equation

The Mistake: Plugging in the wrong number of electrons transferred ($n$) into $E = E^\circ - \frac{0.0591}{n} \log Q$.

How to Avoid:

• $n$ = number of electrons transferred in the balanced overall reaction.
• For $Mg|Mg^{2+}||Cu^{2+}|Cu$, $n = 2$.
• For $Fe|Fe^{2+}||H^+|H_2$, $n = 2$ (since $Fe \rightarrow Fe^{2+} + 2e^-$ and $2H^+ + 2e^- \rightarrow H_2$).

4. Writing the Reaction Quotient $Q$ Incorrectly

The Mistake: Including solids, liquids, or gases with wrong exponents or omitting concentration terms.

How to Avoid:

• Solids and pure liquids have activity = 1 — do not include them in $Q$.
• Gases use partial pressure in bar (not concentration).
• Ions use molar concentration.
• $Q = \frac{[\text{products}]}{[\text{reactants}]}$ (only aqueous ions and gases)
• Example for $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$:
  • $Q = \frac{[Mg^{2+}]}{[Cu^{2+}]}$ (Mg and Cu are solids, so omitted)

5. Confusing $E^\circ_{cell}$ with $E_{cell}$

The Mistake: Using standard reduction potentials directly without adjusting for non-standard conditions.

How to Avoid:

• $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ (standard conditions only)
• $E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q$ (for non-standard concentrations at 298 K)
• Always check: if concentrations are not 1 M, you must use the Nernst equation.

6. Sign Errors in $E^\circ_{cell}$ Calculation

The Mistake: Adding instead of subtracting, or using the wrong sign for the anode potential.

How to Avoid:

• Use the formula: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
• Do not flip the sign of the anode potential — the formula already accounts for it.
• Example: $E^\circ_{Cu^{2+}/Cu} = +0.34\ V$, $E^\circ_{Mg^{2+}/Mg} = -2.37\ V$
  • $E^\circ_{cell} = 0.34 - (-2.37) = +2.71\ V$ (correct)
  • Common mistake: $0.34 + (-2.37) = -2.03\ V$ (wrong)

7. Forgetting to Convert Concentration Units

The Mistake: Using concentrations in units other than molarity (M) without conversion.

How to Avoid:

• All concentrations in the Nernst equation must be in mol/L (M).
• If given in mM, convert: $1\ mM = 0.001\ M$.
• If given in molality (rare), assume it equals molarity for dilute solutions.

8. Omitting the Temperature Factor

The Mistake: Using $\frac{0.0591}{n} \log Q$ at temperatures other than 298 K.

How to Avoid:

• The simplified form $E = E^\circ - \frac{0.0591}{n} \log Q$ is valid only at 298 K. …