Chemistry Q73 – JEE-MAIN 2026

Home Chemistry ElectrochemistryQuestion #388

An electrochemical cell, consist of the following two redox couples,

M^{x +} (a q) / M (s)

[

E_{r e d}^{Θ} = + 0.15 V

] and

F e^{3 +} (a q) / F e (s)

[

E_{r e d}^{Θ} = - 0.036 V

] . The cell EMF (

E_{c e l l}

) is recorded to be

0.2057 V

. If the reaction quotient of the electrochemical reaction is found to be

10^{- 2}

, then the value of

x

is _______ (Nearest integer) [Given : M is a p-block metal and

\frac{2.303 R T}{F} = 0.059 V

]

Enter Calculated Numerical Value:

Clue 1 of 3: Electrochemical Cell Setup

Identify which redox couple acts as the anode and which acts as the cathode based on their standard reduction potentials to determine the standard cell potential.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine Standard Cell Potential (

E_{c e l l}^{Θ}

)✦ Active

The given standard reduction potentials are $E_{r e d}^{Θ} (M^{x +} / M) = + 0.15 V$ and $E_{r e d}^{Θ} (F e^{3 +} / F e) = - 0.036 V$ . Since $M^{x +} / M$ has a higher reduction potential, it acts as the cathode, and $F e^{3 +} / F e$ acts as the anode.

E_{c e l l}^{Θ} = E_{c a t h o d e}^{Θ} - E_{a n o d e}^{Θ} = (+ 0.15 V) - (- 0.036 V) = 0.186 V

Step 2: Apply Nernst Equation to find 'n'○ Expand

The Nernst equation is $E_{c e l l} = E_{c e l l}^{Θ} - \frac{2.303 R T}{n F} \log Q$ . We are given $E_{c e l l} = 0.2057 V$ , $E_{c e l l}^{Θ} = 0.186 V$ , $Q = 10^{- 2}$ , and $\frac{2.303 R T}{F} = 0.059 V$ . Substitute these values into the Nernst equation:

0.2057 = 0.186 - \frac{0.059}{n} \log (10^{- 2})

0.2057 = 0.186 - \frac{0.059}{n} (- 2)

0.2057 - 0.186 = \frac{0.118}{n}

0.0197 = \frac{0.118}{n}

n = \frac{0.118}{0.0197} \approx 5.9898

Rounding to the nearest integer, the number of electrons transferred, $n = 6$ .

💡 Teacher's Secret Hint

Ensure correct sign handling for $\log Q$ and careful arithmetic for $n$ .

Step 3: Determine the value of 'x'○ Expand

The half-reactions for the cell are:

Anode (Oxidation): F e (s) \to F e^{3 +} (a q) + 3 e^{-}

Cathode (Reduction): M^{x +} (a q) + x e^{-} \to M (s)

To balance the electrons in the overall reaction, the number of electrons transferred ( $n$ ) must be the least common multiple of 3 and $x$ . Thus, $n = 3 x$ . Since we found $n = 6$ :

3 x = 6

x = \frac{6}{3}

x = 2

The value of $x$ is 2.

💡 Teacher's Secret Hint

Remember to balance the electrons in the overall cell reaction to correctly determine $n$ in terms of $x$ .

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Chemistry Question 73 – JEE-MAIN 2026

Clue 1 of 3: Electrochemical Cell Setup

🧠 Full Solution Path