Consider the unbalanced chemical equation given below to answer the questions that follow. (i) Write the balanced oxidation half-equation. (ii) Write the balanced reduction half-eq... Consider the unbalanced chemical equation given below to answer the questions that follow. (i) Write the balanced oxidation half-equation. (ii) Write the balanced reduction half-equation. (iii) Deduce the overall balanced equation in acidic medium. Using the bond energy data given below, calculate the enthalpy change for the reaction. A 10ml of 0.020mol L⁻¹ AgNO₃ solution and a 20ml of 0.20mol L⁻¹ CH₃COONa solution are mixed together at 25°C. (i) Determine the moles of Ag⁺ and CH₃COO⁻ present in AgNO₃ solution and CH₃COONa solution respectively. (ii) Determine the concentration of Ag⁺ and CH₃COO⁻ after the two solutions are mixed. (iii) Determine the ionic product (IP) of CH₃COOAg.
Understand the Problem
The question involves balancing chemical equations, calculating enthalpy changes using bond energy data, and determining concentrations and ionic products from a reaction mixture. It covers multiple aspects of chemistry including redox reactions, thermodynamics, and solution chemistry.
Answer
Overall balanced equation: $$ 4 MnO_4^- + 10 H_2O \rightarrow 4 Mn^{2+} + 5 O_2 + 6 H^+ $$ Enthalpy change: $$ \Delta H = -87 \text{ kJ/mol} $$
Answer for screen readers
The overall balanced equation is: $$ 4 MnO_4^- + 10 H_2O \rightarrow 4 Mn^{2+} + 5 O_2 + 6 H^+ $$
The enthalpy change for the reaction $C_2H_4 + Cl_2 \rightarrow C_2H_4Cl_2$ is: $$ \Delta H = -87 \text{ kJ/mol} $$
Steps to Solve
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Writing the Oxidation Half-Equation To find the oxidation half-equation, identify the species that is oxidized. In this case, the manganese in $MnO_4^-$ is reduced. The balanced half-equation for the oxidation of $H_2O$ to $O_2$ in an acidic medium is: $$ 2 H_2O \rightarrow O_2 + 4 H^+ + 4 e^- $$
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Writing the Reduction Half-Equation For the reduction half-equation, we focus on the conversion of $MnO_4^-$ to $Mn^{2+}$ in acidic medium. The balanced half-equation is: $$ MnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4 H_2O $$
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Combining the Half-Equations To derive the overall balanced equation, equalize the electrons in both half-equations. Multiply the oxidation half-equation by 5 and the reduction half-equation by 4 (to equalize the electron transfer): Oxidation: $$ 10 H_2O \rightarrow 5 O_2 + 20 H^+ + 20 e^- $$
Reduction: $$ 4 MnO_4^- + 32 H^+ + 20 e^- \rightarrow 4 Mn^{2+} + 16 H_2O $$
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Final Overall Balanced Equation Add both half-equations and simplify: $$ 4 MnO_4^- + 10 H_2O \rightarrow 4 Mn^{2+} + 5 O_2 + 6 H^+ $$
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Calculating Enthalpy Change Using Bond Energy Identify all bonds broken and formed in the reaction of $C_2H_4 + Cl_2 \rightarrow C_2H_4Cl_2$. Subtract total bond energy of products from total bond energy of reactants.
For example, the bonds involved:
- Bonds broken:
- $ C=C $: 348 kJ/mol
- $ Cl-Cl $: 243 kJ/mol
Total broken = $348 + 243 = 591 \text{ kJ/mol}$
- Bonds formed:
- $ C-Cl $: formed from two $C-Cl$ bonds.
Total formed: $2 * 339 = 678 \text{ kJ/mol}$
- Calculating Final Enthalpy Change The enthalpy change, $\Delta H$, is given by: $$ \Delta H = \text{Bonds broken} - \text{Bonds formed} = 591 - 678 = -87 \text{ kJ/mol} $$
The overall balanced equation is: $$ 4 MnO_4^- + 10 H_2O \rightarrow 4 Mn^{2+} + 5 O_2 + 6 H^+ $$
The enthalpy change for the reaction $C_2H_4 + Cl_2 \rightarrow C_2H_4Cl_2$ is: $$ \Delta H = -87 \text{ kJ/mol} $$
More Information
- The reactions showcase manganese's reduction from $MnO_4^-$ to $Mn^{2+}$ and bond energy calculations which are fundamental concepts in thermodynamics.
- The negative enthalpy change signifies that the reaction is exothermic, releasing energy.
Tips
- Forgetting to balance charges when writing half-equations.
- Miscalculating total bond energies by not accounting for all bonds formed and broken.
- Not simplifying the final overall equation correctly.
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