W redox reactions 13-14.pdf - RHSMartinAPChem - Free Printable

Here are the solutions for the redox reaction worksheet. To solve these, we look at how the oxidation numbers (charges) change for each element.

General Rules to Remember:
* Oxidation is the loss of electrons (oxidation number increases/goes up).
* Reduction is the gain of electrons (oxidation number decreases/goes down).
* The Reducing Agent is the substance that gets oxidized (it gives away electrons).
* The Oxidizing Agent is the substance that gets reduced (it takes electrons).

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1. $Br^-(aq) + MnO_4^-(aq) \rightarrow Br_2(l) + Mn^{2+}(aq)$

Step-by-step Analysis:
* Bromine ($Br$): Starts as $Br^-$ (charge -1) and becomes $Br_2$ (charge 0). The charge goes from -1 to 0. It increased, so Bromine is oxidized.
* Manganese ($Mn$): In $MnO_4^-$, Oxygen is -2. So, $Mn + 4(-2) = -1$, which means $Mn = +7$. It ends as $Mn^{2+}$ (+2). The charge goes from +7 to +2. It decreased, so Manganese is reduced.

Answers:
* What substance is being reduced? $MnO_4^-$ (specifically the Mn inside it)
* What substance is being oxidized? $Br^-$
* What is the reducing agent? $Br^-$
* What is the oxidizing agent? $MnO_4^-$

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2. $MnO_4^- + H_2C_2O_4 \rightarrow CO_2 + Mn^{2+}$

Step-by-step Analysis:
* Manganese ($Mn$): As calculated before, in $MnO_4^-$, Mn is +7. It becomes $Mn^{2+}$ (+2). The charge went down, so it is reduced.
* Carbon ($C$): In oxalic acid ($H_2C_2O_4$), Hydrogen is +1 and Oxygen is -2.
* $2(+1) + 2(C) + 4(-2) = 0$
* $2 + 2C - 8 = 0$
* $2C - 6 = 0 \rightarrow 2C = 6 \rightarrow C = +3$.
* In $CO_2$, Oxygen is -2, so Carbon is +4.
* The charge went from +3 to +4. It went up, so Carbon is oxidized.

Answers:
* What substance is being reduced? $MnO_4^-$
* What substance is being oxidized? $H_2C_2O_4$
* What is the reducing agent? $H_2C_2O_4$
* What is the oxidizing agent? $MnO_4^-$

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3. $As_2O_3 + NO_3^- \rightarrow NO + H_3AsO_4$

Step-by-step Analysis:
* Arsenic ($As$): In $As_2O_3$, Oxygen is -2.
* $2(As) + 3(-2) = 0 \rightarrow 2As = 6 \rightarrow As = +3$.
* In $H_3AsO_4$, Hydrogen is +1 and Oxygen is -2.
* $3(+1) + As + 4(-2) = 0 \rightarrow 3 + As - 8 = 0 \rightarrow As - 5 = 0 \rightarrow As = +5$.
* Charge went from +3 to +5. It went up, so Arsenic is oxidized.
* Nitrogen ($N$): In nitrate ($NO_3^-$), Oxygen is -2.
* $N + 3(-2) = -1 \rightarrow N - 6 = -1 \rightarrow N = +5$.
* In nitrogen monoxide ($NO$), Oxygen is -2, so Nitrogen is +2.
* Charge went from +5 to +2. It went down, so Nitrogen is reduced.

Answers:
* What substance is being reduced? $NO_3^-$
* What substance is being oxidized? $As_2O_3$
* What is the reducing agent? $As_2O_3$
* What is the oxidizing agent? $NO_3^-$

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4. $Cr_2O_7^{2-} + Cl^- \rightarrow Cr^{3+} + Cl_2$

Step-by-step Analysis:
* Chromium ($Cr$): In dichromate ($Cr_2O_7^{2-}$), Oxygen is -2.
* $2(Cr) + 7(-2) = -2 \rightarrow 2Cr - 14 = -2 \rightarrow 2Cr = +12 \rightarrow Cr = +6$.
* It becomes $Cr^{3+}$ (+3).
* Charge went from +6 to +3. It went down, so Chromium is reduced.
* Chlorine ($Cl$): Starts as $Cl^-$ (-1) and becomes $Cl_2$ (0).
* Charge went from -1 to 0. It went up, so Chlorine is oxidized.

Answers:
* What substance is being reduced? $Cr_2O_7^{2-}$
* What substance is being oxidized? $Cl^-$
* What is the reducing agent? $Cl^-$
* What is the oxidizing agent? $Cr_2O_7^{2-}$

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Final Answer:

1.
* Reduced: $MnO_4^-$
* Oxidized: $Br^-$
* Reducing Agent: $Br^-$
* Oxidizing Agent: $MnO_4^-$

2.
* Reduced: $MnO_4^-$
* Oxidized: $H_2C_2O_4$
* Reducing Agent: $H_2C_2O_4$
* Oxidizing Agent: $MnO_4^-$

3.
* Reduced: $NO_3^-$
* Oxidized: $As_2O_3$
* Reducing Agent: $As_2O_3$
* Oxidizing Agent: $NO_3^-$

4.
* Reduced: $Cr_2O_7^{2-}$
* Oxidized: $Cl^-$
* Reducing Agent: $Cl^-$
* Oxidizing Agent: $Cr_2O_7^{2-}$