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Otrada [13]
3 years ago
12

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
Chemistry
1 answer:
romanna [79]3 years ago
3 0

Answer:

Coefficient of H^{+}(aq) is more than 4

Explanation:

Oxidation: Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

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

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

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

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

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

So coefficient of H^{+}(aq) is more than 4

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