An electrochemical cell consist of the following two redox couples, $\mathrm{M}^{\times+}(\mathrm{aq})/\mathrm{M}(\mathrm{s})$ $[\mathrm{E}_{\text{red}}^{\ominus}=+0.15\mathrm{~V}]$ and $\mathrm{Fe}^{3+}(\mathrm{aq})/\mathrm{Fe}(\mathrm{s})$ $[\mathrm{E}_{\text{red}}^{\ominus}=-0.036\mathrm{~V}]$ The cell EMF ($\mathrm{E}_{\text{cell}}$) 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 $\frac{2.303\mathrm{RT}}{\mathrm{~F}}=0.059\mathrm{~V}$]

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Redox Reactions and Electrochemistry

An electrochemical cell consist of the following two redox couples,

M^{\times+} (aq) / M (s)

[E_{red}^{⊖} = + 0.15 V]

and

Fe^{3 +} (aq) / Fe (s)

[E_{red}^{⊖} = - 0.036 V]

The cell EMF (

E_{cell}

) is recorded to be 0.2057 V.
If the reaction quotient of the electrochemical reaction is found to be

1 0^{- 2}

, then the value of x is _____ (Nearest integer)
[Given: M is a p-block metal and

\frac{2.303 RT}{F} = 0.059 V

]

Answer: 2

The couple with the higher reduction potential,

M^{x +} /M (+ 0.15 V)

, is the cathode, and iron is oxidised at the anode, so

E_{cell}^{\circ} = 0.15 - (- 0.036) = 0.186 V

. Balancing electrons between

M^{x +} + x e^{-} \to M

and

Fe \to F e^{3 +} + 3 e^{-}

transfers

n = 3 x

electrons. The Nernst equation gives

E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} lo g Q

, so

0.2057 = 0.186 - \frac{0.059}{3 x} lo g (1 0^{- 2}) = 0.186 + \frac{0.118}{3 x}

. Hence

\frac{0.118}{3 x} = 0.0197

, giving

3 x \approx 6

, so

x = 2

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer