JEE MainChemistry

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+4 marks−1 if incorrectNumericalPrevious-year question

Cell Potential and the Nernst Equation for a Redox Couple

Consider the cell Pt(s)midH_2(g)(1atm)midH^(+)(aq,[H^(+)]=1)∥Fe^(3+)(aq), Fe^(2+)(aq)midPt(s) Given E_(Fe^(3+)/Fe^(2+))°=0.771V and E_(H^(+)/1/2H_2)°=0V, T=298K If the potential of the cell is 0.712V, the ratio of concentration of Fe^(2+) to Fe^(3+) is (Nearest integer)

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Correct answer

The ratio of concentration of ${F e}^{2 +}$ to ${F e}^{3 +}$ is 10.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Cell: $P t (s) ∣ H_{2} (1 atm) ∣ H^{+} (1 M) ∥ {F e}^{3 +}, {F e}^{2 +} ∣ P t (s)$ , $E_{{F e}^{3 +} / {F e}^{2 +}}^{\circ} = 0.771 V$ , $E_{H^{+} / \frac{1}{2} H_{2}}^{\circ} = 0 V$ , $E_{cell} = 0.712 V$ at $T = 298 K$ .

02Find

The ratio of concentration of ${F e}^{2 +}$ to ${F e}^{3 +}$ , i.e., $[{F e}^{2 +}] / [{F e}^{3 +}]$ .

03Strategise

1. Write the half-cell reactions and the net cell reaction: Anode: $\frac{1}{2} H_{2} (g) \to H^{+} (a q) + e^{-}$ Cathode: ${F e}^{3 +} (a q) + e^{-} \to {F e}^{2 +} (a q)$ Overall: $\frac{1}{2} H_{2} (g) + {F e}^{3 +} (a q) \to H^{+} (a q) + {F e}^{2 +} (a q)$ with $n = 1$ . 2. Calculate $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.771 - 0 = 0.771 V$ . 3. Apply the Nernst equation at $298 K$ : $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$ Here, $Q = \frac{[{F e}^{2 +}] [H^{+}]}{[{F e}^{3 +}] P_{H_{2}}^{1 / 2}} = \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]}$ since $[H^{+}] = 1$ and $P_{H_{2}} = 1 atm$ .

04Execute

Substitute the numerical values into the Nernst equation: $0.712 = 0.771 - 0.059 \log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]}$ $0.059 \log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 0.771 - 0.712 = 0.059$ $\log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 1$ $\frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 10^{1} = 10$

✓Verify

Checking the direction of cell potential: $E_{cell} = 0.712 V < E_{cell}^{\circ} = 0.771 V$ , which means $Q > 1$ . Hence, the ratio $[{F e}^{2 +}] / [{F e}^{3 +}] = 10 > 1$ , which is completely consistent.

Hints that build this answer step by step

What is the overall spontaneous cell reaction for this electrochemical cell?
$\frac{1}{2} H_{2} (g) + {F e}^{3 +} (a q) \to H^{+} (a q) + {F e}^{2 +} (a q)$
Which form of the Nernst equation applies to this cell at $298 K$ , where $n = 1$ , $[H^{+}] = 1$ , and $P_{H_{2}} = 1 atm$ ?
$E_{cell} = E_{cell}^{\circ} - 0.0591 \log_{10} (\frac{[{F e}^{2 +}]}{[{F e}^{3 +}]})$
Substitute $E_{cell} = 0.712 V$ and $E_{cell}^{\circ} = 0.771 V$ into the equation to find $\frac{[{F e}^{2 +}]}{[{F e}^{3 +}]}$ . What is the value?
10

✓ 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: 8 September 2026

Quick checks

Students also ask

Why do we use 0.059 instead of 0.0591 or 2.303 RT/F?

At 298 K, 2.303 RT/F evaluates to ~0.0591 V. Here, 0.771 - 0.712 = 0.059 V, which is designed by the examiner to cancel directly with 0.059.

Answer

The ratio of concentration of ${F e}^{2 +}$ to ${F e}^{3 +}$ is 10.

Why each option works or fails

Step-by-step solution

Given: Cell: $P t (s) ∣ H_{2} (1 atm) ∣ H^{+} (1 M) ∥ {F e}^{3 +}, {F e}^{2 +} ∣ P t (s)$ , $E_{{F e}^{3 +} / {F e}^{2 +}}^{\circ} = 0.771 V$ , $E_{H^{+} / \frac{1}{2} H_{2}}^{\circ} = 0 V$ , $E_{cell} = 0.712 V$ at $T = 298 K$ .
Find: The ratio of concentration of ${F e}^{2 +}$ to ${F e}^{3 +}$ , i.e., $[{F e}^{2 +}] / [{F e}^{3 +}]$ .
Strategise: 1. Write the half-cell reactions and the net cell reaction: Anode: $\frac{1}{2} H_{2} (g) \to H^{+} (a q) + e^{-}$ Cathode: ${F e}^{3 +} (a q) + e^{-} \to {F e}^{2 +} (a q)$ Overall: $\frac{1}{2} H_{2} (g) + {F e}^{3 +} (a q) \to H^{+} (a q) + {F e}^{2 +} (a q)$ with $n = 1$ . 2. Calculate $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.771 - 0 = 0.771 V$ . 3. Apply the Nernst equation at $298 K$ : $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$ Here, $Q = \frac{[{F e}^{2 +}] [H^{+}]}{[{F e}^{3 +}] P_{H_{2}}^{1 / 2}} = \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]}$ since $[H^{+}] = 1$ and $P_{H_{2}} = 1 atm$ .
Execute: Substitute the numerical values into the Nernst equation: $0.712 = 0.771 - 0.059 \log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]}$ $0.059 \log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 0.771 - 0.712 = 0.059$ $\log \frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 1$ $\frac{[{F e}^{2 +}]}{[{F e}^{3 +}]} = 10^{1} = 10$
Verify: Checking the direction of cell potential: $E_{cell} = 0.712 V < E_{cell}^{\circ} = 0.771 V$ , which means $Q > 1$ . Hence, the ratio $[{F e}^{2 +}] / [{F e}^{3 +}] = 10 > 1$ , which is completely consistent.

Cell Potential and the Nernst Equation for a Redox Couple

What feels right?

The ratio of concentration of ${F e}^{2 +}$ to ${F e}^{3 +}$ is 10.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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