Question

3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)+NOg) (acidic) a. HCO (a)+Ags)+NH3(ag) (basic) b. H2CO(a)+Ag(NHs)2 (aq)

Answer 1

Explanation:

(a)   The given reaction equation is as follows.

        $Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

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.

      $Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b)   The given reaction equation is as follows.

        $HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

      $4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

      $H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$