I think you forgot one. Thank you. **Redox Reaction Electron Transfer Problem****Problem Statement:**For the following reaction, determine the number of electrons transferred. If the reaction is not a redox reaction, please enter "0" for your answer.\[ \text{3Cu(s)} + \text{8HNO}_3\text{(aq)} \rightarrow \text{3CuSO}_4\text{(aq)} + \text{8NO(g)} + \text{4H}_2\text{O(l)} \]**Answer:**\[ \_\_\_\_\_\_\_\_\_ \text{electrons} \]---**Instructions on How to Solve:**1. **Identify Oxidation and Reduction:** - Determine which elements are oxidized and reduced by assigning oxidation states to each element in the reaction.2. **Calculate Electrons Transferred:** - Calculate the total number of electrons gained and lost using the change in oxidation states.3. **Enter the Result:** - Insert the total number of electrons transferred into the input box. If no electrons are transferred, enter "0."

Consider the change in oxidation number of the species involved in the reaction as the reactants get converted to products. Divide the reaction into oxidation-half reaction and reduction-half reaction.In the given reaction, S has an oxidation state of -2 in CuS and +6 in CuSO4. The change in oxidation number is 6-(-2) =8. An increase in oxidation number indicates that S has undergone oxidation. So, the oxidation-half reaction is 3CuS(s) →3CuSO4 +24 e- we have 24 electrons in the above half reaction because there are three S atoms and each S atoms has lost 8 electrons.In the given reaction, N has an oxidation state of +5 in HNO3 and +2 in NO The change in oxidation number is 2-(5) =-3. An decrease in oxidation number indicates that N has undergone reduction. So, the Reduction-half reaction is 8HNO3(aq) +24 e-→8NO(g) we have 24 electrons in the above half reaction because there are Eight N atoms and each N atoms has gained 3 electrons and undergone reduction. So, 24 electrons are transferred from the S of CuS to N of the HNO3. Answer is 24