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Write the following redox reactions usin...

Write the following redox reactions using half equations:
a. `Zn(s)+PbCl_(2)(aq)rarrPb(s)+ZnCl_(2)(aq)`.
b. `2Fe^(3+)(aq)+2I^(ө)(aq)rarrI_(2)(aq)+2Fe^(2+)(aq)`
c. `2Na(s)+Cl_(2)(g)rarr2NaCl(s)`
d. `Mg(s)+Cl_(2)(g)rarrMgCl_(2)(s)`
e. `Zn(s)+2H^(o+)(aq)rarrZn^(2+)(aq)+H_(2)(g)` .

Text Solution

Verified by Experts

(i) `Zn(s) rarr Zn^(2+)+2e^(-) ("oxidation") , Pb^(2+)(aq)+2e^(-) rarrPb(s)("reduction")`
Zn is oxidised to `Zn^(2+), Pb^(2+) ("reduction") , Pb^(2+)` is the oxidiser and Zn is the reducer
(ii) `2Fe^(3+)+2e^(-) rarr 2Fe^(2+) ("reduction") , 2I^(-)rarrI_(2)+2e^(-)("oxidation")`
`Fe^(3+)` reduced to `Fe^(2+) I^(-)` is oxidiased to `I_(2) , I^(-)` is the reducer and `Fe^(3+)` is the oxidiser
(iii)`2Na rarr2Na^(+)+2e^(-)("oxidation") , CI_(2)+2e^(-)rarr2CI^(-)`(reduction)
Na is oxidised to `Na^(+) "and" CI_(2)` is reduced to `CI^(-) , CI_(2)+2e^(-) rarr 2CI^(-)` (reduction )
Mg is oxidised o `Zn^(2+)` while `H^(+)` is reduced to `H_(2)` , Zn is ht reducer and `H^(+)` is the oxidiser
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