A) 22.4 mL
B) 112 mL
C) 168 mL
D) 336 mL
Correct Answer: C
Solution :
| [c] At cathode |
| \[{{H}_{2}}O+2{{e}^{-}}\xrightarrow{\,}\,{{H}_{2}}+2O{{H}^{\odot -}}\] |
| At Anode |
| \[{{H}_{2}}O\xrightarrow{\,}\frac{1}{2}\,{{O}_{2}}+2{{H}^{\oplus }}+2{{e}^{-}}\] |
| i.e., on passing 2 mol of current, 1 mol \[{{H}_{2}}(g)\] and \[\frac{1}{2}\,mol\,\,{{O}_{2}}(g)\] is liberated at cathode and anode respectively. |
| From \[Q=it\] |
| \[=\,1\times (16\times 60+5)=965\,\,C\] |
| the amount of gases liberated by passing 965 C of electricity \[=\frac{1.5\times 965}{2\times 96500}\text{mol}\] |
| = 0.0075 mol |
| and value of gases liberated |
| \[=0.0075\times 22400\,mL=168\,mL\] |
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