A current of 0.75 A is passed through an acidic solution of CuSO4 for 10 minutes. The volume of oxygen liberated at anode (at STP) will be

Question

A current of 0.75 A is passed through an acidic solution of CuSO4 for 10 minutes. The volume of oxygen liberated at anode (at STP) will be (A) 0.261 dm3 (B) 0.261 cm3 (C) 0.261 × 102mL (D) 0.261m3

Tardigrade Admin · Accepted Answer

2O2- → O2 +4e Mole of e =(0.75×10×60/96500) Mole of O2=(4.66×10-3/4)=0.0261 L

Q. A current of $0.75$ A is passed through an acidic solution of $C u S O_{4}$ for 10 minutes. The volume of oxygen liberated at anode (at STP) will be

$0.261 d m^{3}$

$0.261 c m^{3}$

$0.261 \times 1 0^{2} m L$

$0.261 m^{3}$

$2 O^{2 -} \to O_{2} + 4 e$
Mole of e $= \frac{0.75 \times 10 \times 60}{96500}$
Mole of $O_{2} = \frac{4.66 \times 1 0 ^{- 3}}{4} = 0.0261 L$

Q. A current of 0.75 A is passed through an acidic solution of CuSO4​ for 10 minutes. The volume of oxygen liberated at anode (at STP) will be

0.261dm3

0.261cm3

0.261×102mL

0.261m3