Q.19E - Chemistry (OpenStax)

Question

Predict the products of the following reactions and balance the equations.

(a) Zn is added to a solution of Cr₂(SO₄)₃ in acid.

(b) FeCl₂ is added to a solution containing an excess of Cr₂O₇²⁻ in hydrochloric acid.

(c) Cr²⁺ is added to Cr₂O₇²⁻ in acid solution.

(d) Mn is heated with CrO₃.

(e) CrO is added to 2HNO₃ in water.

(f) FeCl3 is added to an aqueous solution of NaOH.

Step-by-Step Solution

Answer

The balanced reactions are as follow:

 

a) \[Zn(s) + C{r_2}{\left( {S{O_4}} \right)_3}(aq) + 2{H_3}{O^ + }(aq) \to ZnS{O_4}(aq) + 2CrS{O_4}(aq) + {H_2}(g) + 2{H_2}O(l)\]

b) \[6FeC{l_2}(s) + C{r_2}O_7^{2 - }(aq) + 14HCl(aq) \to 6FeC{l_3}(aq) + 2CrC{l_3}(aq) + 2C{l^ - }(aq) + 7{H_2}O(l)\]

c) \[6C{r^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 8C{r^{3 + }} + 7{H_2}O\]

d) \[9Mn(s) + 8Cr{O_3}(s) \to 9M{n_3}{O_4}(s) + 4C{r_2}{O_3}(s)\]

e) \[CrO(s) + 2NO_3^ - (aq) + 3{H_3}{O^ + }(aq) \to C{r^{2 + }}(aq) + 2NO_3^ - (aq) + 2{H_2}O(l)\]

f) \[FeC{l_3}(s) + 3NaOH(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3C{l^ - }(aq)\]

Step 1: Redox Reaction

Redox is a type of chemical reaction in which the oxidation states of atoms are changed. Redox reactions are characterized by the actual or formal transfer of electrons between chemical species.

Step 2: Balancing the given equations

The following equations are balanced in following manner:

 

a)

\(Zn(s) + C{r_2}{\left( {S{O_4}} \right)_3}(aq) + {H_3}{O^ + }(aq) \to ZnS{O_4}(aq) + CrS{O_4}(aq) + {H_2}(g)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:Zn \to Z{n^{2 + }} + 2{e^ - }\\O:2C{r^{3 + }} \to 2C{r^{6 + }} + 6{e^ - }\\R:2{S^{6 + }} + 8{e^ - } \to 2{S^{2 + }}\\\overline {Zn + 2C{r^{3 + }} + 2{S^{6 + }} \to Z{n^{2 + }} + 2C{r^{6 + }} + 2{S^{2 + }}} \end{aligned}\)

 

From, that we have,

\(Zn(s) + C{r_2}{\left( {S{O_4}} \right)_3}(aq) + 2{H_3}{O^ + }(aq) \to ZnS{O_4}(aq) + 2CrS{O_4}(aq) + {H_2}(g) + 2{H_2}O(l)\)

 

b)

\(FeC{l_2}(s) + C{r_2}O_7^{2 - }(aq) + HCl(aq) \to FeC{l_3}(aq) + CrC{l_3}(aq) + C{l^ - }(aq) + {H_2}O(l)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:6F{e^{2 + }} \to 6F{e^{3 + }} + 6{e^ - }\\R:C{r_2}O_7^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\\\overline {6F{e^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O} \end{aligned}\)

 

From, that we have,

\(6FeC{l_2}(s) + C{r_2}O_7^{2 - }(aq) + 14HCl(aq) \to 6FeC{l_3}(aq) + 2CrC{l_3}(aq) + 2C{l^ - }(aq) + 7{H_2}O(l)\)

 

c)

\(C{r^{2 + }}(aq) + C{r_2}O_7^{2 - }(aq) + {H^ + }(aq) \to C{r^{3 + }}(aq) + {H_2}O(l)\)

 

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:6C{r^{2 + }} \to 6C{r^{3 + }} + 6{e^ - }\\R:C{r_2}O_7^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\\\overline {6C{r^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 8C{r^{3 + }} + 7{H_2}O}\end{aligned}\)

From, that we have,

\(6C{r^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 8C{r^{3 + }} + 7{H_2}O\)

 

d)

\(Mn(s) + Cr{O_3}(s) \to M{n_3}{O_4}(s) + C{r_2}{O_3}(s)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:9Mn \to 9M{n^{\frac{8}{3} + }} + 24{e^ - }\\R:8C{r_2}{O_3} + 24{e^ - } \to 4C{r_2}{O_3}\\\overline {9Mn + 8C{r_2}{O_3} \to 9M{n^{3 + }} + 4C{r_2}{O_3}} \end{aligned}\)

 

From, that we have,

\(9Mn(s) + 8Cr{O_3}(s) \to 9M{n_3}{O_4}(s) + 4C{r_2}{O_3}(s)\)

 

 

e)

\(CrO(s) + 2NO_3^ - (aq) + {H_3}{O^ + }(aq) \to C{r^{2 + }}(aq) + 2NO_3^ - (aq) + {H_2}O(l)\)

After balancing the oxidation number and reduction number we have:

\(CrO(s) + 2NO_3^ - (aq) + 3{H_3}{O^ + }(aq) \to C{r^{2 + }}(aq) + 2NO_3^ - (aq) + 2{H_2}O(l)\)

 

f)

\(FeC{l_3}(s) + NaOH(aq) \to Fe{(OH)_3}(s) + N{a^ + }(aq) + C{l^ - }(aq)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:3Cl \to 3Cl + 3{e^ - }\\R:3N{a^ + } + 3{e^ - } \to 3Na\\\overline {6F{e^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O} \end{aligned}\)

 

From, that we have,

\(FeC{l_3}(s) + 3NaOH(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3C{l^ - }(aq)\)

 

 

 

Hence, the final answer is,

The balanced reactions are as follow:

 

a) \(Zn(s) + C{r_2}{\left( {S{O_4}} \right)_3}(aq) + 2{H_3}{O^ + }(aq) \to ZnS{O_4}(aq) + 2CrS{O_4}(aq) + {H_2}(g) + 2{H_2}O(l)\)

b) \(6FeC{l_2}(s) + C{r_2}O_7^{2 - }(aq) + 14HCl(aq) \to 6FeC{l_3}(aq) + 2CrC{l_3}(aq) + 2C{l^ - }(aq) + 7{H_2}O(l)\)

c) \(6C{r^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 8C{r^{3 + }} + 7{H_2}O\)

d) \(9Mn(s) + 8Cr{O_3}(s) \to 9M{n_3}{O_4}(s) + 4C{r_2}{O_3}(s)\)

e) \(CrO(s) + 2NO_3^ - (aq) + 3{H_3}{O^ + }(aq) \to C{r^{2 + }}(aq) + 2NO_3^ - (aq) + 2{H_2}O(l)\)

f) \(FeC{l_3}(s) + 3NaOH(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3C{l^ - }(aq)\)