JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

An electrochemical cell is constructed using half cells in the direction of spontaneous change

$F e (O H)_{2} (s) + 2 e^{-} \to F e (s) + 2 O H^{-} (a q) E^{0} = - 0.88$ V

and $A g B r (s) + e^{-} \to A g (s) + B r^{-} (a q) E^{0} = + 0.07$ V

Which of the following option is correct ?

A
Overall reaction $F e (s) + 2 O H^{-} (a q) + 2 A g B r (s) ⇌ F e (O H)_{2} (s) + 2 A g (s) + 2 B r^{-} (a q)$
B
$E_{ce l l}^{0} = - 0.95$ V
C
Fe is reduced in the electrochemical cell
D
$E_{ce l l}^{0}$ is an extensive property

Solution

For a spontaneous cell, the half cell with higher reduction potential is the cathode. Here $A g B r / A g$ ( $+ 0.07$ V) is the cathode and $F e (O H)_{2} / F e$ ( $- 0.88$ V) is the anode (Fe gets oxidised).

$E_{ce l l}^{0} = E_{c a t h o d e}^{0} - E_{an o d e}^{0} = 0.07 - (- 0.88) = + 0.95$ V, which is positive (so option 2 is wrong, and Fe is oxidised, not reduced).

Cell potential is intensive (option 4 wrong).

Combining the half-reactions gives the overall reaction in option (1):

$F e (s) + 2 O H^{-} (a q) + 2 A g B r (s) ⇌ F e (O H)_{2} (s) + 2 A g (s) + 2 B r^{-} (a q)$

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