Chapter 19, Problem 1SAQ

Interpretation Introduction

Interpretation: The correct coefficients for each reactant and product are to be chosen for the following reaction from the given options:

$\_V{O}_2^{+}\left(aq\right)+\_\_Sn\left(s\right)+\_\_H+\left(aq\right)\to \_\_V{O}^{2+}\left(aq\right)+\_\_S{n}^{2+}\left(aq\right)+\_\_H_{2}O(l)$

a) 2, 1, 4 → 2, 1, 2

b) 1, 1, 2 → 1, 1, 1

c) 2, 1, 2 → 2, 1, 1

d) 2, 1, 2 → 2, 1, 2

Answer to Problem 1SAQ

Solution:

To solve this problem, begin by assigning oxidation states to all atoms and identify the substances being oxidized and reduced.

Explanation of Solution

The equation for this reaction as follows:

$V{O}_2^{+}\left(aq\right)+Sn\left(s\right)+H^{+}\left(aq\right)\to V{O}^{2+}\left(aq\right)+S{n}^{2+}\left(aq\right)+H_{2}O\left(l\right)$

Separate the overall reaction into two half-reactions: one for oxidation and other for reduction.

$\begin{array}{l}Oxidation:Sn\left(s\right)\to S{n}^{2+}\left(aq\right)\\ \mathrm{Reduction}:V{O}_2^{+}\left(aq\right)\to V{O}^{2+}\left(aq\right)\end{array}$

Balance each half-reaction; balance O by adding $H_{2}O$ and balance H by adding $H^{+}.$

$\begin{array}{l}Sn\left(s\right)\to S{n}^{2+}\left(aq\right)\\ V{O}_2^{+}\left(aq\right)\to V{O}^{2+}\left(aq\right)+H_{2}O\left(l\right)\\ 2H^{+}\left(aq\right)+V{O}_2^{+}\left(aq\right)\to V{O}^{2+}\left(aq\right)+H_{2}O\left(l\right)\end{array}$

Balance each-half reaction with respect to charge by adding electrons. Make the sum of the charges on both sides of the equation equal by adding as many electrons as necessary.

$\begin{array}{l}Sn\left(s\right)\to S{n}^{2+}\left(aq\right)+2e^{-}\\ 2H^{+}\left(aq\right)+V{O}_2^{+}\left(aq\right)+e^{-}\to V{O}^{2+}\left(aq\right)+H_{2}O\left(l\right)\end{array}$

Make the number of electrons in both-half reactions equal by multiplying one or both half-reactions by a small whole number.

$\begin{array}{l}Sn\left(s\right)\to S{n}^{2+}\left(aq\right)+2e^{-}\\ 2\left[2H^{+}\left(aq\right)+V{O}_2^{+}\left(aq\right)+e^{-}\to V{O}^{2+}\left(aq\right)+H_{2}O\left(l\right)\right]\\ 4H^{+}\left(aq\right)+2V{O}_2^{+}\left(aq\right)+2e^{-}\to 2V{O}^{2+}\left(aq\right)+2H_{2}O\left(l\right)\end{array}$

Add the two half-reaction together, cancelling electrons and other species as necessary.

$\begin{array}{l}Sn\left(s\right)\to S{n}^{2+}\left(aq\right)+\cancel{2e^{-}}\\ 4H^{+}\left(aq\right)+2V{O}_2^{+}\left(aq\right)+\cancel{2e^{-}}\to V{O}^{2+}\left(aq\right)+2H_{2}O\left(l\right)\\ 2V{O}_2^{+}\left(aq\right)+Sn\left(s\right)+4H^{+}\left(aq\right)\to 2V{O}^{2+}\left(aq\right)+S{n}^{2+}\left(aq\right)+2H_{2}O\left(l\right)\end{array}$

Conclusion:

Thus, the coefficient of each reactant and product will be $2,1,4\to 2,1,2$ and the correct answer will be option (a).