Respuesta :
Explanation:
(a) The given reaction equation is as follows.
[tex]Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)[/tex] (acidic)
So, here the reduction and oxidation-half reactions will be as follows.
Oxidation-half reaction: [tex]Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}[/tex]
Reduction-half-reaction: [tex]NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)[/tex]
As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.
[tex]Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)[/tex] (acidic)
(b) The given reaction equation is as follows.
[tex]HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)[/tex] (basic)
So, here the reduction and oxidation-half reactions will be as follows.
Reduction-half reaction: [tex]HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)[/tex]
Oxidation-half reaction: [tex]Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}[/tex]
Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.
[tex]4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-} [/tex]
Therefore, balancing the whole reaction equation in the basic medium as follows.
[tex]H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)[/tex]
