Question

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

Answer 1

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