1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# In the electrolysis of H2O, 11.2 litre of H2 liberated at cathode at NTP. How much O2 will be liberated at anode under the same conditions?

A
11.2 litre
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
22.4 litre
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
32g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5.6 litre
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 5.6 litreH2OL→H2(g)+12O2(g) ↓ ↓ ↓1 mole 1 mole 1/2 mole As we know from PV=nRT P→ Pressure V→ Volume of gas n→ moles of gas R→ universal gas constant T→ Temperaturefrom PV=nRT V∝nSo, H2OL→H2(g)+12O2(g) Volm(lt)Volm2(lt)Therefore, volume of O2 litrated at anode =Volm2(∴Volm=11.2lt) =11.22=5.6lt

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Volume of Gases and Number of Moles
CHEMISTRY
Watch in App
Join BYJU'S Learning Program