The balanced equation for a spontaneous reaction which involves the tin and zinc redox couple using the following standard reduction potentials in acid solution is: $\begin{array}{ll} \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathrm{E}^{\circ}=+1.23 \mathrm{~V} \ \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+} & \mathrm{E}^{\circ}=+0.13 \mathrm{~V} \ \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{~s}) & \mathrm{E}^{\circ}=-0.76 \mathrm{~V} \end{array}$ Zn + Sn⁴⁺ $\rarr$ Zn²⁺ + Sn²⁺

Question

Quizplus · Accepted Answer

The balanced equation can be determined by comparing the reduction potentials for the two half-reactions and ensuring that the electrons cancel out. The reduction potentials are:

Zn2+(aq) + 2e- -> Zn(s) E° = -0.763 V
Sn2+(aq) + 2e- -> Sn(s) E° = -0.136 V

Since zinc has a more negative reduction potential, it is the reducing agent and will undergo oxidation. Tin will be reduced. The balanced equation is:

Zn(s) + Sn2+(aq) -> Zn2+(aq) + Sn(s)

Adding sulfuric acid to both sides of the equation gives the acid solution:

Zn(s) + Sn2+(aq) + 2H+(aq) -> Zn2+(aq) + Sn(s) + H2(g)

Finally, the sulfur ions must be balanced to give the overall equation:

Zn(s) + SnS2(aq) + 2H+(aq) -> ZnS(s) + Sn(s) + H2(g)

The Balanced Equation for a Spontaneous Reaction Which Involves the Tin