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
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Correct Answer: Option 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}
\(Sn ^{2+}+ Cr _{2} O _{7}^{2-}+ H ^{-} \rightarrow Sn ^{4+}+ Cr ^{3+}+ H _{2} O\)
\(Sn ^{2+} \rightarrow Sn ^{4+}+2 e ^{-}\)(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}