Question

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Fill in all blanks with numbers so if the term is not in the equation make it 0. CrO42- (aq) + N2O(g)+ H+(aq) + OH-(aq) + H2O(l) LaTeX: \longrightarrow⟶ Cr3+(aq) + NO(g) + H+(aq) + OH-(aq) + H2O(l) oxidizing agent is (enter just the formula of the species, e.g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-) reducing agent is (enter just the formula of the species, e.g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-)

Answer 1

Answer:

$2(CrO_4)^{2-} (aq) + 3N_2O(g)+ 10H^+(aq) + \longrightarrow 2Cr^{3+}(aq) + 6NO(g) + 5H_2O(l)$

Reducing agent: nitrogen.

Oxidizing agent: chromium.

Explanation:

Hello,

In this case, the first step is to rewrite the undergoing chemical reaction indicating the oxidation state for each atom as shown below:

$(Cr^{+6}O^{-2}_4)^{2-} (aq) + N^+_2O^{-2}(g)+ H^+(aq) + OH^-(aq) + H_2O(l) \longrightarrow Cr^{3+}(aq) + N^{+2}O^{-2}(g) + H^+(aq) + OH^-(aq) + H_2O(l)$

In such a way, we observe the following half-reactions:

* Reduction as chromium undergoes a decrease in its oxidation state:

$(Cr^{+6}O_4^{-2})^{2-}+8H^++3e^-\rightarrow Cr^{+3}+4H_2O$

* Oxidation as nitrogen undergoes an increase in its oxidation state:

$N^+_2O^{-2}+H_2O\rightarrow 2N^{+2}O^{-2}+2H^++2e^-$

In such a way, we state that the nitrogen is the reducing agent as it is oxidized and chromium the oxidizing agent as it is reduced, therefore, the balanced chemical reaction turns out:

$2(CrO_4)^{2-} (aq) + 3N_2O(g)+ 10H^+(aq) + \longrightarrow 2Cr^{3+}(aq) + 6NO(g) + 5H_2O(l)$

Best regards.