1. Answer the questions below according to the following unbalanced redox reacti
ID: 629636 • Letter: 1
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1. Answer the questions below according to the following unbalanced redox reaction In3+ + Li(s) --------> In (s) + Li+ a. Write the balanced half reaction for the species that is being oxidized b. Write the balanced half reaction for the species that is being reduced c. Write the balanced redox reaction d. Which species is the reducing agent in this redox reaction? e. Which species is the oxidizing agent in this redox reaction? 2. Answer the questions below according to the following unbalanced redox reaction AsH3(g) + BrO4- (aq) As(s) + BrO3- (aq) (Acidic Solution) a. Write the balanced half reaction for the species that is at the anode b. Write the balanced half reaction for the species that is at the cathode c. Write the balanced redox reaction d. Which species is the reducing agent in this redox reaction? e. Which species is the oxidizing agent in this redox reaction? 3. Write the balanced redox reaction for the same unbalanced redox reaction in #1 but this time in a basic solution AsH3(g) + BrO4- (aq) As(s) + BrO3- (aq) (Basic Solution)Explanation / Answer
Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS(s) ---> Cu2+(aq) + SO42-(aq) And the unbalanced half-reaction for NO3- is: NO3-(aq) --> NO(g) nsert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Balance oxygen by adding H2O to one side of each half-reaction. CuS + 4 H2O ---> Cu2+ + SO42- NO3- --> NO + 2 H2O Add H3O+ to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H2O to the other side. For basic solutions: add H2O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH- to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+ NO3- + 4 H3O+ --> NO + 6 H2O Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+ + 8 e- Reduction: NO3- + 4 H3O+ + 3 e- --> NO + 6 H2O
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