Consider the cell When the potential of the cell is 0.712 V at 298 K, the ratio is ______ (Nearest Integer). Given: [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Consider the cell

${Pt}_{(s)} | H_{2} (g,1 atm) H^{+} (aq, 1 M) ‖ {Fe}^{3 +} (a q) . {Fe}^{2 +} (a q) | Pt (s)$

When the potential of the cell is 0.712 V at 298 K, the ratio $\frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]}$ is ______ (Nearest Integer).

Given: ${Fe}^{3 +} + e^{-} = {Fe}^{2 +}, E^{0} {Fe}^{3 +}, {Fe}^{2 +} | Pt = 0.771$

$\frac{2.303 RT}{F} = 0.06 V$ [2023]

Ans.
(10)

$Anode ⇒ \frac{1}{2} H_{2} (g) \to H^{+} (a q) + e^{-}$

$Cathode ⇒ F e^{3 +} + e^{-} \to F e^{2 +}$

$Overall: \frac{1}{2} H_{2} + F e^{3 +} \overset{n = 1}{\to} H^{+} + F e^{2 +}$

$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{1} \log \frac{[F e^{2 +}]}{[F e^{3 +}]} \times \frac{[H^{+}]}{{[P_{H_{2}}]}^{1 / 2}}$

$0.712 = 0.771 - 0.059 log \frac{[F e^{2 +}]}{[F e^{3 +}]}$

$\log \frac{[F e^{2 +}]}{[F e^{3 +}]} = 1$

So $\frac{[F e^{2 +}]}{[F e^{3 +}]} = 10$

Want Us to Email you About Special Offers & Updates?