Penyetaraan reaksi molekul
CONTOH 01\(\ce{Cd}_{\text{(s)}} + \ce{NiO2}_{\text{(s)}} + \ce{H2O}_{\text{(l)}} \rightarrow \ce{Cd(OH)2}_{\text{(aq)}} + \ce{Ni(OH)2}_{\text{(aq)}}\)
\(\ce{\underset{\color{red} 0}{Cd}}_{\text{(s)}} + \ce{\underset{\color{red}+4}{Ni}O2}_{\text{(s)}} + \ce{H2O}_{\text{(l)}} \rightarrow \ce{\underset{\color{red} +2}{Cd}(OH)2}_{\text{(aq)}} + \ce{\underset{\color{red}+2}{Ni}(OH)2}_{\text{(aq)}}\)
Oksidasi = Biloks Cd berubah dari 0 menjadi +2 (naik 2) × 1
Reduksi = Biloks Ni berubah dari +4 menjadi +2 (turun 2) × 1
\(\color{blue}1 \color{black}\cdot \ce{Cd}_{\text{(s)}} + \color{blue}1 \color{black}\cdot\ce{NiO2}_{\text{(s)}} + \ce{H2O}_{\text{(l)}} \rightarrow \color{blue}1 \color{black}\cdot\ce{Cd(OH)2}_{\text{(aq)}} + \color{blue}1 \color{black}\cdot\ce{Ni(OH)2}_{\text{(aq)}}\)
Selanjutnya setarakan jumlah atom H
\(\color{blue}1 \color{black}\cdot \ce{Cd}_{\text{(s)}} + \color{blue}1 \color{black}\cdot\ce{NiO2}_{\text{(s)}} + \color{blue}2 \color{black}\cdot\ce{H2O}_{\text{(l)}} \rightarrow \color{blue}1 \color{black}\cdot\ce{Cd(OH)2}_{\text{(aq)}} + \color{blue}1 \color{black}\cdot\ce{Ni(OH)2}_{\text{(aq)}}\)
Hasil akhir:
$$\bbox[yellow, 5px, border: 2px solid red] {\ce{Cd}_{\text{(s)}} + \ce{NiO2}_{\text{(s)}} + 2\ce{H2O}_{\text{(l)}} \rightarrow \ce{Cd(OH)2}_{\text{(aq)}} + \ce{Ni(OH)2}_{\text{(aq)}}}$$
CONTOH 02
\(\ce{KMnO4}_{\text{(aq)}} + \ce{Na2SO3}_{\text{(aq)}} + \ce{H2SO4}_{\text{(aq)}} \rightarrow \ce{K2SO4}_{\text{(aq)}} + \ce{MnSO4}_{\text{(aq)}} + \ce{Na2SO4}_{\text{(aq)}} + \ce{H2O}_{\text{(l)}}\)
\(\ce{K\underset{\color{red}+7}{Mn}O4}_{\text{(aq)}} + \ce{Na2\underset{\color{red}+4}{S}O3}_{\text{(aq)}} + \ce{H2SO4}_{\text{(aq)}} \rightarrow \ce{K2SO4}_{\text{(aq)}} + \ce{\underset{\color{red}+2}{Mn}SO4}_{\text{(aq)}} + \ce{Na2\underset{\color{red}+6}{S}O4}_{\text{(aq)}} + \ce{H2O}_{\text{(l)}}\)
Oksidasi = Biloks S berubah dari +4 menjadi +6 (naik 2) × 5
Reduksi = Biloks Mn berubah dari +7 menjadi +2 (turun 5) × 2
\(\color{blue}2 \color{black}\cdot\ce{KMnO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO3}_{\text{(aq)}} + \ce{H2SO4}_{\text{(aq)}} \rightarrow \color{blue}1 \color{black}\cdot\ce{K2SO4}_{\text{(aq)}} + \color{blue}2 \color{black}\cdot\ce{MnSO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO4}_{\text{(aq)}} + \ce{H2O}_{\text{(l)}}\)
Setarakan jumlah atom S
\(\color{blue}2 \color{black}\cdot\ce{KMnO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO3}_{\text{(aq)}} + \color{blue}3 \color{black}\cdot\ce{H2SO4}_{\text{(aq)}} \rightarrow \color{blue}1 \color{black}\cdot\ce{K2SO4}_{\text{(aq)}} + \color{blue}2 \color{black}\cdot\ce{MnSO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO4}_{\text{(aq)}} + \ce{H2O}_{\text{(l)}}\)
Setarakan jumlah atom H
\(\color{blue}2 \color{black}\cdot\ce{KMnO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO3}_{\text{(aq)}} + \color{blue}3 \color{black}\cdot\ce{H2SO4}_{\text{(aq)}} \rightarrow \color{blue}1 \color{black}\cdot\ce{K2SO4}_{\text{(aq)}} + \color{blue}2 \color{black}\cdot\ce{MnSO4}_{\text{(aq)}} + \color{blue}5 \color{black}\cdot\ce{Na2SO4}_{\text{(aq)}} + \color{blue}3\color{black}\cdot\ce{H2O}_{\text{(l)}}\)
Hasil akhir:
$$\bbox[yellow, 5px, border: 2px solid red] {2\ce{KMnO4}_{\text{(aq)}} + 5\ce{Na2SO3}_{\text{(aq)}} +3\ce{H2SO4}_{\text{(aq)}} \rightarrow \ce{K2SO4}_{\text{(aq)}} + 2\ce{MnSO4}_{\text{(aq)}} + 5\ce{Na2SO4}_{\text{(aq)}} + 3\ce{H2O}_{\text{(l)}}}$$
CONTOH 03
\(\ce{ZnS}_{\text{(s)}} + \ce{HNO3}_{\text{(aq)}} \rightarrow \ce{ZnSO4}_{\text{(aq)}} + \ce{NO}_{\text{(g)}} + \ce{H2O}_{\text{(l)}}\)
\(\ce{Zn}\ce{\underset{\color{red}-2}{S}} + \ce{H\underset{\color{red}+5}{N}O3} \rightarrow \ce{Zn\underset{\color{red}+6}{S}O4} + \ce{\underset{\color{red}+2}{N}O} + \ce{H2O}\)
Oksidasi = Biloks S berubah dari −2 menjadi +6 (naik 8) × 3
Reduksi = Biloks N berubah dari +5 menjadi +2 (turun 3) × 8
\(\color{blue}3 \color{black}\cdot\ce{ZnS}_{\text{(s)}} + \color{blue}8 \color{black}\cdot\ce{HNO3}_{\text{(aq)}} \rightarrow \color{blue}3 \color{black}\cdot\ce{ZnSO4}_{\text{(aq)}} + \color{blue}8 \color{black}\cdot\ce{NO}_{\text{(g)}} + \ce{H2O}_{\text{(l)}}\)
Setarakan jumlah atom H
\(\color{blue}3 \color{black}\cdot\ce{ZnS}_{\text{(s)}} + \color{blue}8 \color{black}\cdot\ce{HNO3}_{\text{(aq)}} \rightarrow \color{blue}3 \color{black}\cdot\ce{ZnSO4}_{\text{(aq)}} + \color{blue}8 \color{black}\cdot\ce{NO}_{\text{(g)}} + \color{blue}4 \color{black}\cdot\ce{H2O}_{\text{(l)}}\)
Hasil akhir:
$$\bbox[yellow, 5px, border: 2px solid red] {3\ce{ZnS}_{\text{(s)}} +8\ce{HNO3}_{\text{(aq)}} \rightarrow 3\ce{ZnSO4}_{\text{(aq)}} + 8\ce{NO}_{\text{(g)}} + 4\ce{H2O}_{\text{(l)}}}$$
CONTOH 04
\(\ce{FeS} + \ce{HNO3} \rightarrow \ce{Fe(NO3)3} + \ce{Fe2(SO4)3} + \ce{NO} + \ce{H2O}\) suasana asam
Setarakan jumlah atom Fe, S, dan N
\(3\ce{FeS} + 4\ce{HNO3} \rightarrow \ce{Fe(NO3)3} + \ce{Fe2(SO4)3} + \ce{NO} + \ce{H2O}\)
Tulis dalam reaksi ion
\(3\ce{Fe}^{2+} + 3\ce{S}^{2-} + 4\ce{H}^{+} +\cancelto{\ce{NO3}^{-}}{ 4\ce{NO3}^{-}} \rightarrow \ce{Fe}^{3+} + \cancel{3\ce{NO3}^{-}} + 2\ce{Fe}^{3+} + 3\ce{SO4}^{2-} + \ce{NO} + \ce{H2O}\)
\(3\ce{\underset{\color{red}+6}{Fe}}^{2+} + 3\ce{\underset{\color{red}-6}{S}}^{2-} + 4\ce{H}^{+} + \ce{\underset{\color{red}+5}{N}O3}^{-} \rightarrow 3\ce{\underset{\color{red}+9}{Fe}}^{3+} + 3\ce{\underset{\color{red}+18}{S}O4}^{2-} + \ce{\underset{\color{red}+2}{N}O} + \ce{H2O}\)
Reaksi Oksidasi:
Biloks Fe berubah dari +6 menjadi +9 (naik 3)
Biloks S berubah dari −6 menjadi +18 (naik 24)
Total naik 27 × 1
Reaksi Reduksi:
Biloks N berubah dari +5 menjadi +2 (turun 3) × 9
\(\color{blue}3 \color{black} \cdot \ce{Fe}^{2+} + \color{blue}3 \color{black} \cdot\ce{S}^{2-} + \color{blue}4 \color{black} \cdot\ce{H}^{+} + \color{blue}9 \color{black} \cdot\ce{NO3}^{-} \rightarrow \color{blue}3 \color{black} \cdot\ce{Fe}^{3+} + \color{blue}3 \color{black} \cdot\ce{SO4}^{2-} + \color{blue}9 \color{black} \cdot\ce{NO} + \ce{H2O}\)
Menyetarakan jumlah muatan
Jumlah muatan sebelah kiri = 3(+2) + 3(−2) + 4(+1) + 9(−1) = −5
Jumlah muatan sebelah kanan = 3(+3) + 3(−2) + 9(0) + 0 = +3
Karena berada pada suasana asam, maka tambahkan \(\ce{H}^{+}\) sebanyak selisih jumlah muatan pada bagian yang kelebihan elektron
\(\color{blue}3 \color{black} \cdot \ce{Fe}^{2+} + \color{blue}3 \color{black} \cdot\ce{S}^{2-} + \color{blue}4 \color{black} \cdot\ce{H}^{+} + \color{blue}9 \color{black} \cdot\ce{NO3}^{-} + \color{blue}8 \color{black}\cdot\ce{H}^{+}\rightarrow \color{blue}3 \color{black} \cdot\ce{Fe}^{3+} + \color{blue}3 \color{black} \cdot\ce{SO4}^{2-} + \color{blue}9 \color{black} \cdot\ce{NO} + \ce{H2O}\)
Selanjutnya setarakan jumlah atom H dengan menambahkan \(5\ce{H2O}\) di ruas kanan
\(\color{blue}3 \color{black} \cdot \ce{Fe}^{2+} + \color{blue}3 \color{black} \cdot\ce{S}^{2-} + \color{blue}4 \color{black} \cdot\ce{H}^{+} + \color{blue}9 \color{black} \cdot\ce{NO3}^{-} + \color{blue}8 \color{black}\cdot\ce{H}^{+}\rightarrow \color{blue}3 \color{black} \cdot\ce{Fe}^{3+} + \color{blue}3 \color{black} \cdot\ce{SO4}^{2-} + \color{blue}9 \color{black} \cdot\ce{NO} + \ce{H2O} +\color{blue}5 \color{black} \cdot\ce{H2O}\)
\(\color{blue}3 \color{black} \cdot \ce{Fe}^{2+} + \color{blue}3 \color{black} \cdot\ce{S}^{2-} + \color{blue}12 \color{black} \cdot\ce{H}^{+} + \color{blue}9 \color{black} \cdot\ce{NO3}^{-} \rightarrow \color{blue}3 \color{black} \cdot\ce{Fe}^{3+} + \color{blue}3 \color{black} \cdot\ce{SO4}^{2-} + \color{blue}9 \color{black} \cdot\ce{NO} +\color{blue}6\color{black} \cdot\ce{H2O}\)
\(\color{blue}3 \color{black} \cdot \ce{Fe}^{2+} + \color{blue}3 \color{black} \cdot\ce{S}^{2-} + \color{blue}12 \color{black} \cdot\ce{H}^{+} + \color{blue}12 \color{black} \cdot\ce{NO3}^{-} \rightarrow \color{blue}1 \color{black} \cdot\ce{Fe}^{3+} + + \color{blue}3 \color{black} \cdot\ce{NO3}^{-} + \color{blue}2\color{black} \cdot\ce{Fe}^{3+}+\color{blue}3 \color{black} \cdot\ce{SO4}^{2-} + \color{blue}9 \color{black} \cdot\ce{NO} +\color{blue}6\color{black} \cdot\ce{H2O}\)
Selanjutnya tulis kembali dalam bentuk reaksi molekulnya.
Hasil akhir:
$$\bbox[yellow, 5px, border: 2px solid red] {3\ce{FeS} + 12\ce{HNO3} \rightarrow \ce{Fe(NO3)3} + \ce{Fe2(SO4)3} + 9\ce{NO} + 6\ce{H2O}}$$