A) 2, 1, 2
B) 1, 2, 2
C) 2, 2, 1
D) None of these
Correct Answer: B
Solution :
| [b] (i) The half reaction for oxidation is, \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\xrightarrow[{}]{{}}C{{O}_{2}}\] |
| Balancing carbon atoms on both sides, \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\xrightarrow[{}]{{}}2C{{O}_{2}}\] |
| Balancing hydrogen atoms on both sides, \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\xrightarrow[{}]{{}}2C{{O}_{2}}+2{{H}^{+}}\] |
| Balancing the charge on both sides, \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\xrightarrow[{}]{{}}2C{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\](balanced) |
| (ii) The half reaction for reduction is- \[{{H}_{2}}{{O}_{2}}\xrightarrow[{}]{{}}{{H}_{2}}O\] |
| Balancing oxygen atoms on both sides, \[{{H}_{2}}{{O}_{2}}\xrightarrow[{}]{{}}2{{H}_{2}}O\] |
| Balancing hydrogen atoms, \[{{H}_{2}}{{O}_{2}}+2{{H}^{+}}\xrightarrow[{}]{{}}2{{H}_{2}}O\] |
| Balancing the charge, \[{{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\xrightarrow[{}]{{}}2{{H}_{2}}O\] (balanced) Now, adding both equation, |
| \[{{H}_{2}}{{C}_{2}}{{O}_{4}}+{{H}_{2}}{{O}_{2}}\xrightarrow[{}]{{}}2C{{O}_{2}}+2{{H}_{2}}O\] |
| This is balanced equation. |
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