A 1M solution of H2SO4 is electrolyzed. Select right statement with respect to concentration of electrolyte and products at anode and cathode respectively. Given 2SO42−→S2O82−+2e−;Eo=−2.01VH2O(l)→2H(aq) ++12O2(g)+2e−;Eo=−1.23V

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A 1M solution of H₂SO₄ is electrolyzed. Select right statement with respect to concentration of electrolyte and products at anode and cathode respectively. Given
$\begin{array}{l} 2 {SO}_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 e^{-}; E^{o} = - 2.01 V \\ H_{2} O_{(l)} \to 2 H_{(aq)}^{+} + \frac{1}{2} O_{2 (g)} + 2 e^{-}; E^{o} = - 1.23 V \end{array}$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

concentration of $H_{2} {SO}_{4}$ increases ; $O_{2}, H_{2}$

concentration of $H_{2} {SO}_{4}$ remain constant $H_{2}, O_{2}$

concentration of $H_{2} {SO}_{4}$ remains constant $S_{2} O_{8}^{2 -}, H_{2}$

concentration of $H_{2} {SO}_{4}$ decreases: $O_{2}, H_{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$C a t h o d e : 2 H^{+} + 2 e^{-} \to H_{2} . A s E^{o} o f S O_{4}^{2 -} a n d S_{2} O_{8}^{2 -} l e s s t h a n H_{2} O / O_{2} . S o t h e r e a c t i o s a t a n o d e a r e H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} .$