**Title: Unbalancing Redox Reactions: An Educational Exploration****Introduction**Consider the following unbalanced redox reaction:\[ \text{SCN}^-(\text{aq}) + \text{OCl}^-(\text{aq}) \rightleftharpoons \text{SO}_4^{2-}(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) + \text{NO}_3^-(\text{aq}) + \text{Cl}^-(\text{aq}) \]**Discussion Points****a. Oxidation Numbers Inquiry:**- **Sulfur in \(\text{SO}_4^{2-}\):** What is the oxidation number for the sulfur atom?- **Carbon in \(\text{CO}_3^{2-}\):** What is the oxidation number for the carbon atom?- **Nitrogen in \(\text{NO}_3^-\):** What is the oxidation number for the nitrogen atom?**b. Relative Oxidation Numbers:**Given that the carbon atom and nitrogen atom in \(\text{SCN}^-\) are arbitrarily assigned the same oxidation numbers as found in \(\text{CO}_3^{2-}\) and \(\text{NO}_3^-\), respectively, determine the necessary oxidation number for the sulfur atom in \(\text{SCN}^-\).**c. Balancing Challenge:**Balance the redox reaction in a basic solution.**Conclusion**This exercise encourages a deeper understanding of oxidation numbers and redox balancing, essential skills in chemistry education. Explore each question thoroughly to grasp the concepts of electron transfer and oxidation states.**Related Resources**- Oxidation Number Rules- Redox Reactions in Basic Solutions- Practical Exercises in Redox Balancing**End of Section**

Name: **Title: Unbalancing Redox Reactions: An Educational Exploration****I
Uploaded: 2022-12-02T03:55:46.000Z
Channel: Bartleby
Description: **Title: Unbalancing Redox Reactions: An Educational Exploration****Introduction**Consider the following unbalanced redox reaction:\[ \text{SCN}^-(\text{aq}) + \text{OCl}^-(\text{aq}) \rightleftharpoons \text{SO}_4^{2-}(\text{aq}) + \text{CO}_3^{2-}

The oxidation number of an element is the charge which an atom of the element carries in its ion or appears to have when exists in the combined state with one or more than one atom. Oxidation numbers are also termed as oxidation state.a. We can suppose the oxidation number of sulfur be x in SO42-. Therefore, x + 4(-2) = -2 x - 8 = -2 x = -2+8 x = +6 Thus, the oxidation number of sulfur is+6 in SO42-. We can suppose the oxidation number of carbon be y in CO32-. Therefore, y + 3(-2) = -2 y - 6 = -2 y = -2+6 y = +4 Thus, the oxidation number of carbon is +4 in CO32-. We can suppose the oxidation number of nitrogen be z in NO3-. Therefore, z + 3(-2) = -1 z - 6 = -1 z = -1+6 z = +5 Thus, the oxidation number of nitrogen is +5 in NO3-.b. Now, if we take the oxidation number of C-atom and N-atom as +4 and +5 respectively in SCN-. Then the value for the oxidation number of S-atom in SCN- is calculated as follows. We can suppose the oxidation number of sulfur be x in SCN-. Therefore, x + 4 + 5 = -1 x + 9 = -1 x = -1+9 x = +8 Thus, the oxidation number of sulfur is +8 in SCN-.c. The first step is to write the two half-reactions. Oxidation half-reaction: SCN-→SO42-+CO32-+NO3- Reduction half-reaction: OCl-→Cl- As we can see, that atoms other than H and O are already balanced. Therefore, the second step is to balance the oxygen on either side of the half-reactions by adding H2O molecules to the deficient side. Oxidation half-reaction: SCN-+10H2O→SO42-+CO32-+NO3- Reduction half-reaction: OCl-→Cl-+H2O Now, the third step is to balance the H-atoms. Oxidation half-reaction: SCN-+10H2O+20OH-→SO42-+CO32-+NO3-+20H2O Reduction half-reaction: OCl-+2H2O→Cl-+H2O+2OH- The fourth step is to balance the charge on either side of the half-reactions by adding electrons. Oxidation half-reaction: SCN-+10H2O+20OH-→SO42-+CO32-+NO3-+20H2O+16e- Reduction half-reaction: OCl-+2H2O+2e-→Cl-+H2O+2OH- Now, to make the electron change equivalent in both the reaction, for the same the reduction half-reaction is multiplied by 8. Then we will add both the half reactions. The addiction result is shown below. SCN-+10H2O+20OH- → SO42-+CO32-+NO3-+20H2O+16e-+8OCl-+16H2O+16e- +8Cl-+8H2O+16OH- After simplification of the above reaction, we get a balanced reaction in the basic medium. SCN-+8OCl-+4OH- → SO42-+CO32-+NO3-+8Cl-+2H2O