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Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient for the H+(aq) ion.

Cr2O72–(aq) + Sn2+(aq) → Cr3+(aq) + Sn4+(aq)

(A) 1 (no coefficient written)
(B) 2
(C) 3
(D) 4
(E) More than 4

Respuesta :

Answer:

Coefficient of [tex]H^{+}(aq)[/tex] is more than 4

Explanation:

Oxidation: [tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)[/tex]

  • Balance charge: [tex]Sn^{2+}(aq)-2e^{-}\rightarrow Sn^{4+}(aq)[/tex]......(1)

Reduction: [tex]Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)[/tex]

  • Balance Cr: [tex]Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)[/tex]
  • Balance O and H in acidic medium: [tex]Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)[/tex]
  • Balance charge: [tex]Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)[/tex].......(2)

[tex][3\times Equation-(1)]+Equation(2)[/tex] gives balanced equation:

[tex]3Sn^{2+}(aq)+Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 3Sn^{4+}(aq)+2Cr^{3+}(aq)+7H_{2}O(l)[/tex]

So coefficient of [tex]H^{+}(aq)[/tex] is more than 4

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