This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Step 1: Recall the definitions of oxidation and reduction. Oxidation is the loss of electrons, resulting in an increase in oxidation state. Reduction is the gain of electrons, resulting in a decrease in oxidation state. Step 2: Analyze each half-reaction to determine if electrons are gained or lost, and how the oxidation state changes. (i)* Fe^3+ + 3e^- Fe Fe^3+ gains 3 electrons to become Fe. The oxidation state changes from +3 to 0. This is a gain of electrons, so it is reduction. (ii)* Cr Cr^3+ + 3e^- Cr loses 3 electrons to become Cr^3+. The oxidation state changes from 0 to +3. This is a loss of electrons, so it is oxidation. (iii)* MnO_4^2- MnO_4^- + e^- In MnO_4^2-, the oxidation state of Mn is +6 (x + 4(-2) = -2 x = +6). In MnO_4^-, the oxidation state of Mn is +7 (y + 4(-2) = -1 y = +7). MnO_4^2- loses 1 electron to become MnO_4^-. The oxidation state of Mn changes from +6 to +7. This is a loss of electrons, so it is oxidation. (iv)* Li^+ + e^- Li Li^+ gains 1 electron to become Li. The oxidation state changes from +1 to 0. This is a gain of electrons, so it is reduction. Here are the classifications: (i) Fe^3+ + 3e^- Fe: Reduction (ii) Cr Cr^3+ + 3e^-: Oxidation (iii) MnO_4^2- MnO_4^- + e^-: Oxidation (iv) Li^+ + e^- Li: Reduction That's 2 down. 3 left today — send the next one.