Calculate Sum of Water Moles and Electrons Involved in Reaction

To find the value of $(x + y)$, we need to write the balanced redox reaction for the reaction between hydrogen sulfide ($\text{H}_2\text{S}$) and acidified potassium permanganate ($\text{KMnO}_4$).

1. Oxidation Half-Reaction

In an acidic medium, $\text{H}_2\text{S}$ is oxidized to elemental sulfur ($\text{S}$): $\text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2e^-$

For $5\text{ moles}$ of $\text{H}_2\text{S}$, multiplying the equation by $5$: $5\text{H}_2\text{S} \rightarrow 5\text{S} + 10\text{H}^+ + 10e^-$

From this half-reaction, the total number of moles of electrons transferred during the oxidation of $5\text{ moles}$ of $\text{H}_2\text{S}$ is: $y = 10$

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2. Reduction Half-Reaction

Permanganate ion ($\text{MnO}_4^-$) is reduced to $\text{Mn}^{2+}$ in acidic medium: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

To consume the $10\text{ moles}$ of electrons released by $5\text{ moles}$ of $\text{H}_2\text{S}$, we multiply the reduction half-reaction by $2$: $2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

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3. Overall Balanced Ionic Equation

Combining the oxidation and reduction half-reactions gives: $2\text{MnO}_4^- + 5\text{H}_2\text{S} + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 8\text{H}_2\text{O}$

From the balanced equation:

• The number of moles of water ($\text{H}_2\text{O}$) produced when $5\text{ moles}$ of $\text{H}_2\text{S}$ reacts completely is $x = 8$.
• The number of moles of electrons involved in the reaction is $y = 10$.

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4. Calculation

$x + y = 8 + 10 = 18$