During electrolysis of `H_2SO_4`(aq) with high charge density, `H_2S_2O_8` formed as by product. In such electrolysis 22.4L `H_2(g)` and 8.4 L `O_2(g)` liberated at 1 atm and 273 K at electrode. The moles of `H_2S_2O_8` formed is :
A
0.25
B
0.5
C
0.75
D
1
Text Solution
Verified by Experts
The correct Answer is:
A
Topper's Solved these Questions
ELECTROCHEMISTRY
NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|1 Videos
ELECTROCHEMISTRY
NARENDRA AWASTHI|Exercise Level 1 (Q.91 To Q.120)|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
During electrolysis of one mole or aq HCOOK . How many moles of H_2 is formed overall ?
At T(K), the ratio of kinetic energies of 4g of H_(2) (g) and 8g of O_(2) (g) is
Oxidation state of S in H_(2)SO_(5) and H_(2)S_(2)O_(8) respectively are
2H_(2(g)) + O_(2(g)) rarr 2H_(2)O_((l)) , DeltaH= -ve and DeltaG= -ve . Then the reaction is
Oxidation sttae of S in H_(2)SO _(5) and H_(2) S _(2) O_(8) respectively are
Calculate the partial pressures of O_2 and H_2 in a mixture of 3 moles of O_2 and 1 mole of H_2 at S.T.P.
H_(2)SO_(4(aq)) + 2KOH_((aq)) rarr K_(2)SO_(4(aq)) + 2H_(2)O_((l)), Delta H for the above reaction is
The reaction between H_2O_2 and KMnO_4 is 2KMnO_4 + 3H_2SO_4 + 5H_2O_2 to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2 . In a reaction excess of H_2O_2 is added to 0.1 mole of acidified KMnO_4 solution. Then the volume of 02 gas liberated at STP is