Redox Reactions And Electrochemistry Question 126
Question: A current being passed for two hour through a solution of an acid liberating 11.2 litre of oxygen at NTP at anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same time [BVP 2003]
Options:
A) 16 g
B) 63 g
C) 31.5 g
D) 8 g
Show Answer
Answer:
Correct Answer: B
Solution:
At STP, 1 mole of oxygen gas =32g=22.4L
Therefore,11.2L of $O_2=\frac{32}{22.4}×11.2=16g$
Therefore,$\frac{M_{Cu}}{M_{O_2}}=\frac{E_{Cu}}{E_O}$
$⇒\frac{M_{Cu}}{16}=(\frac{63}{2})(\frac{16}{2})$
$⇒M_Cu=63g$
BETA
