JEE Main 2026 · 2 Apr · Shift 1 · Chemistry Q57

← JEE Main 2026 · 2 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 57 of 75

An electrochemical cell is constructed using half cells in the direction of spontaneous change
$\ce{Fe(OH)2(s) + 2e- -> Fe(s) + 2OH-(aq)}\qquad E^\theta=-0.88\ \mathrm V$
and $\ce{AgBr(s) + e- -> Ag(s) + Br-(aq)}\qquad E^\theta=+0.07\ \mathrm V$
Which of the following option is correct ?

Overall reaction

\ce{Fe(s) + 2OH-(aq) + 2AgBr(s) <=> Fe(OH)2(s) + 2Ag(s) + 2Br-(aq)}

E^\theta_\text{cell}=-0.95

Fe is reduced in the electrochemical cell

E^\theta_\text{cell}

is an extensive property

Answer and worked solution

Official answer: A

Spontaneous direction. The couple with lower $E^\theta$ (Fe, −0.88 V) is oxidised and AgBr (+0.07 V) is reduced:
$\ce{Fe + 2OH- + 2AgBr -> Fe(OH)2 + 2Ag + 2Br-}.$

Other options.

$E^\theta_\text{cell}=0.07-(-0.88)=+0.95$ V, not −0.95 V.
Fe is oxidised, not reduced.
$E^\theta$ is intensive.

(A)

Topics:Galvanic cells and EMF

← Previous question Next question →

JEE Main 2026 · 2 Apr · Shift 1, Chemistry question 57