Answer:
(i)
In the reaction of dihydrogen with metals to form metal hydrides, it acts as an
oxidising agent.
\[2\,Na\,(s)+{{H}_{2}}(g)\xrightarrow{Heat}\,2\,N{{a}^{+}}{{H}^{-}}(s)\]
Here, Na has been oxidised to \[N{{a}^{+}}\]
while dihydrogen has been reduced to hydride \[({{H}^{-}})\] ion.
(ii) In the reaction of heated cupric oxide
with dihydrogen to form \[{{H}_{2}}O\] and copper metal, dihydrogen acts as a
reducing agent.
\[CuO(s)+{{H}_{2}}(s)\xrightarrow{Heat}\,Cu(s)+{{H}_{2}}O(g)\]
Here, CuO is reduced to Cu while dihydrogen
is oxidised to \[{{H}_{2}}O\]
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