JEE MainChemistry

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Determining an Ion Concentration Ratio Using the Nernst Equation

Q: Why is n = 2 in the Nernst equation?

Because the cathode reaction is M^(3+) + 2e^(-) → M^(+) and the anode reaction is H_2 → 2H^(+) + 2e^(-), exactly 2 moles of electrons are transferred per mole of reaction.

Pt(s)|H_2(g)(1bar)||H^(+)(aq)(1M)∥M^(3+)(aq), M^(+)(aq)|Pt(s) The E_cell for the given cell is 0.1115V at 298K when ([M^(+)(aq)])/([M^(3+)(aq)]) = 10^a The value of a is Given: E^0M^(3+)/M^(+) = 0.2V 2.303RT/F = 0.059V

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Correct answer

The value of $a$ is 3.

Option analysis

Why each option works or fails

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Solution

StepWorking

01Given

Cell: $P t (s) ∣ H_{2} (g) (1 bar) ∣ H^{+} (a q) (1 M) ∥ M^{3 +} (a q), M^{+} (a q) ∣ P t (s)$ . $E_{cell} = 0.1115 V$ at $298 K$ . $\frac{[M^{+}]}{[M^{3 +}]} = 10^{a}$ . $E_{M^{3 +} / M^{+}}^{\circ} = 0.2 V$ , $\frac{2.303 R T}{F} = 0.059 V$ .

02Find

Find the value of the integer/exponent $a$ .

03Strategise

1. Write the half-cell reactions and overall cell reaction: Anode (oxidation): $H_{2} (g) ⟶ 2 H^{+} (aq) + 2 e^{-}$ , with $E^{\circ} = 0.0 V$ . Cathode (reduction): $M^{3 +} (aq) + 2 e^{-} ⟶ M^{+} (aq)$ , with $E^{\circ} = 0.2 V$ . Number of electrons transferred, $n = 2$ . Overall: $H_{2} (g) + M^{3 +} (aq) ⟶ 2 H^{+} (aq) + M^{+} (aq)$ . $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.2 - 0 = 0.2 V$ . 2. Set up the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log Q$ , where $Q = \frac{[M^{+}] [H^{+}]^{2}}{[M^{3 +}] \cdot P_{H_{2}}} = \frac{[M^{+}]}{[M^{3 +}]} = 10^{a}$ . 3. Solve for $a$ .

04Execute

$0.1115 = 0.2 - \frac{0.059}{2} \log (10^{a})$ $\frac{0.059}{2} \cdot a = 0.2 - 0.1115 = 0.0885$ $a = \frac{2 \times 0.0885}{0.059} = \frac{0.1770}{0.059} = 3$

✓Verify

Check: $\frac{0.059}{2} \times 3 = 0.0885$ . $0.2 - 0.0885 = 0.1115 V$ , which exactly matches $E_{cell}$ .

Hints that build this answer step by step

What is the overall cell reaction and the number of electrons transferred ( $n$ )?
$H_{2} (g) + M^{3 +} (a q) \to 2 H^{+} (a q) + M^{+} (a q)$ with $n = 2$
Given that the standard hydrogen electrode has $E^{0} = 0 V$ , what is the standard cell potential $E_{c e l l}^{0}$ and the form of the reaction quotient $Q$ ?
$E_{c e l l}^{0} = 0.2 V$ and $Q = \frac{[H^{+}]^{2} [M^{+}]}{p_{H_{2}} [M^{3 +}]} = \frac{[M^{+}]}{[M^{3 +}]}$
Substitute the given values into the Nernst equation $E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} \log Q$ to find $a$ . What is $a$ ?
$a = 3$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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Students also ask

Why is n = 2 in the Nernst equation?

Because the cathode reaction is $M^{3 +} + 2 e^{-} \to M^{+}$ and the anode reaction is $H_{2} \to 2 H^{+} + 2 e^{-}$ , exactly 2 moles of electrons are transferred per mole of reaction.

Answer

The value of $a$ is 3.

Why each option works or fails

Step-by-step solution

Given: Cell: $P t (s) ∣ H_{2} (g) (1 bar) ∣ H^{+} (a q) (1 M) ∥ M^{3 +} (a q), M^{+} (a q) ∣ P t (s)$ . $E_{cell} = 0.1115 V$ at $298 K$ . $\frac{[M^{+}]}{[M^{3 +}]} = 10^{a}$ . $E_{M^{3 +} / M^{+}}^{\circ} = 0.2 V$ , $\frac{2.303 R T}{F} = 0.059 V$ .
Find: Find the value of the integer/exponent $a$ .
Strategise: 1. Write the half-cell reactions and overall cell reaction: Anode (oxidation): $H_{2} (g) ⟶ 2 H^{+} (aq) + 2 e^{-}$ , with $E^{\circ} = 0.0 V$ . Cathode (reduction): $M^{3 +} (aq) + 2 e^{-} ⟶ M^{+} (aq)$ , with $E^{\circ} = 0.2 V$ . Number of electrons transferred, $n = 2$ . Overall: $H_{2} (g) + M^{3 +} (aq) ⟶ 2 H^{+} (aq) + M^{+} (aq)$ . $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.2 - 0 = 0.2 V$ . 2. Set up the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log Q$ , where $Q = \frac{[M^{+}] [H^{+}]^{2}}{[M^{3 +}] \cdot P_{H_{2}}} = \frac{[M^{+}]}{[M^{3 +}]} = 10^{a}$ . 3. Solve for $a$ .
Execute: $0.1115 = 0.2 - \frac{0.059}{2} \log (10^{a})$ $\frac{0.059}{2} \cdot a = 0.2 - 0.1115 = 0.0885$ $a = \frac{2 \times 0.0885}{0.059} = \frac{0.1770}{0.059} = 3$
Verify: Check: $\frac{0.059}{2} \times 3 = 0.0885$ . $0.2 - 0.0885 = 0.1115 V$ , which exactly matches $E_{cell}$ .

Determining an Ion Concentration Ratio Using the Nernst Equation

What feels right?

The value of $a$ is 3.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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