The balanced equation for the reaction of tin(II) salt with potassium heptaoxodichloromate (VI) in an acidic medium can be represented as \(SSn ^{2+}+ fCr _{2} 2 O _{7}^{2-}+ gH ^{+} \rightarrow hSn ^{4+}+ i -\) \(Cr ^{3+}+ jH _{2} O .\) e, \(f , g , i\) and \(j\) respectively
A. \(3,5,6,1\) and \(4\) B. \(3,1,14,3,2\) and 7 C. \(3,2,6,1,5\) and 6 D. \(5,2,1,5,3\) and 2
Correct Answer: B
Explanation
\(Sn ^{2+}+ Cr _{2} O _{7}^{2-}+ H ^{-} \rightarrow Sn ^{4+}+ Cr ^{3+}+ H _{2} O\) \(Sn ^{2+} \rightarrow Sn ^{4+}+2 e ^{-}\)(oxidation) \(Cr _{2} O ^{2-}+14 H ^{-}+6 e ^{-} \rightarrow 2 Cr ^{1-}+7 H _{2} O\) \begin{aligned} &\text { Multiply the oxidation half cell by } 3\\ &\left[\frac{\left(\begin{array}{l} 3 Sn ^{2+} \rightarrow 3 Sn ^{4+}+6 e ^{-}+ Cr _{2} O _{7}^{2-} \\ +14 H ^{+}+6 e \rightarrow 2 Cr ^{3+}+7 H _{2} O \end{array}\right)}{\left(\begin{array}{c} 3 Sn ^{2+}+ Cr _{2} O _{7}+14 H \\ \rightarrow 3 Sn ^{4+}+2 Cr ^{3+}+7 H _{2} O \end{array}\right)}\right]\\ &e=3, f=1, g=14, h=3 \text {, }\\ &i=2, j=7 \end{aligned}
