The balanced equation for the reaction of Tin (ii) salt with potassium heptaoxodichromate (VI) in...
The balanced equation for the reaction of Tin (ii) salt with potassium heptaoxodichromate (VI) in an acidic medium can be represented as \(eSn^{2+} + fCr_2O_7^{2+} + gH^+ \rightarrow hSn^{4+} + iCr^{3+} + jH_2O\). e, f, g, h, i, and j are respectively:
- A) 3, 5, 6, 3, 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) 3, 1, 14, 3, 2 and 7
Explanation
To determine the coefficients e, f, g, h, i, and j, we need to balance the given chemical equation:
\(eSn^{2+} + fCr_2O_7^{2-} + gH^+ \rightarrow hSn^{4+} + iCr^{3+} + jH_2O\).
First, let's balance the tin (Sn) atoms. Since tin is present in the form of Sn2+ on the left side and Sn4+ on the right side, to balance the tin atoms, we need 3 Sn2+ on the left and 3 Sn4+ on the right. So, e = 3 and h = 3:
\(3Sn^{2+} + fCr_2O_7^{2-} + gH^+ \rightarrow 3Sn^{4+} + iCr^{3+} + jH_2O\).Next, let's balance the chromium (Cr) atoms. Since Cr is present in the form of Cr2O72- on the left side and Cr3+ on the right side, to balance the chromium atoms, we need 1 Cr2O72- on the left and 2 Cr3+ on the right. So, f = 1 and i = 2:
\(3Sn^{2+} + Cr_2O_7^{2-} + gH^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + jH_2O\).Now, we need to balance the oxygen (O) atoms. There are 7 O atoms on the left side in the Cr2O72- ion. On the right side, O atoms are present only in the water (H2O) molecules. So, to balance the O atoms, we need 7 H2O molecules. Therefore, j = 7:
\(3Sn^{2+} + Cr_2O_7^{2-} + gH^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O\).Finally, to balance the hydrogen (H) atoms and charge, we need 14 H+ ions on the left side. So, g = 14:
\(3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O\).Thus, the balanced equation has the coefficients e = 3, f = 1, g = 14, h = 3, i = 2, and j = 7, which corresponds toOption B.