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Correct Answer – A
a. `H_(2)S` is oxidised because a more electronegative element `Cl_(2)` is added to `H`. Alternatively, a more electropositive element `H` has been removed from S. `Cl_(2)` is reduced due to addition of `H` to it.
b. `Al` is oxidised because oxygen is added to it. `Fe_(3)O_(4)` (ferrous ferric oxide) is reduced beacuse oxygen has been removed from it.
c. `2overset(0)Na(s)+overset(0)H_(2)(g)rarr2overset(0)Naoverset(-1)H(s)`
`2Naimplies2oveset(+1)Na+2e^(-)`(oxidation)
`H_(2)+2e^(-)rarr2H^(ө)`(reduction)
Therefore, `Na` is oxidised to the concept of electronegativity, the more electronegative element `H` is added to `Na`, so `Na` is oxidised and `H_(2)` is reduced.
(i) `H_(2)S` is oxidised because a more electronegative element , chlorine is added to hydrogen (or a more electropositive element , hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it .
(ii) Aluminium is oxidised because oxygen is added to it . ferrous ferric oxide `(Fe_(3)O_(4))` is reduced because oxygen has beem removed from it .
(iii) With the careful application of the concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced .
Reaction (iii) chosen here prompts us to think in terms of another way to define redox reactions .