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23.6: Multimedia Resources for Chapter 23

Difficulty Level: At Grade Created by: CK-12

Copy and distribute the lesson worksheets. Ask students to complete the worksheets alone or in pairs as a review of lesson content.

Balancing Redox Equations Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

Steps in the balancing redox equations process.

  1. Determine the oxidation number for all atoms in the reaction.
  2. Determine which atom is being oxidized and which is being reduced.
  3. Write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.
  4. Write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.
  5. If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.
  6. Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.
  7. Balance all other atoms in each half-reaction except \begin{align*}H\end{align*} and \begin{align*}O\end{align*}.
  8. Balance the \begin{align*}H\end{align*} and \begin{align*}O\end{align*} according to either (a) or (b) depending on whether the reaction is acidic or basic.
    1. If the reaction is acidic, add \begin{align*}H_2O\end{align*} and \begin{align*}H^+\end{align*}. Balance \begin{align*}O\end{align*} first by adding \begin{align*}H_2O\end{align*}, then balance \begin{align*}H\end{align*} by adding \begin{align*}H^+\end{align*}. Charge should now be balanced.
    2. If the reaction is basic, add \begin{align*}OH^-\end{align*} and \begin{align*}H_2O\end{align*}. Balance charge first by adding \begin{align*}OH^-\end{align*}, then balance \begin{align*}O\end{align*} by adding \begin{align*}H_2O\end{align*}. The \begin{align*}H\end{align*} should now be balanced.
  9. Once the half-reactions are balanced, find the lowest common multiple (LCM) for the electrons in the two half-reactions.
  10. Multiply each half-reaction by a whole number so that the total number of electrons in the reduction half-reaction equals the total number of electrons in the oxidation half-reaction, and they each equal the LCM.
  11. Add the two half-reactions and cancel those species that are common to both sides.
  12. Check the equation to be sure that it is balanced by both atoms and charge.

Example of an acidic redox reaction balancing.

Given skeleton: \begin{align*}MnO_4^- + C_2O_4^{2-} \rightarrow Mn^{2+}+ CO_2 \qquad \text{(in acid)}\end{align*}

Step 1:

\begin{align*}& \underset{{\color{red}{+7}} \ \ {\color{red}-2} \ \ }{MnO^-_4} && + && \underset{{\color{red}+3} \ {\color{red}-2} \ \ \ }{C_2O^{2-}_4} && \rightarrow && \underset{{\color{red}+2} \quad \ }{Mn^{2+}} && + && \underset{{\color{red}+4} \ {\color{red}-2} \ \ }{CO_2} && \text{(in acid solution)}\end{align*}

Step 2: \begin{align*}Mn^{+7}\end{align*} is being reduced to \begin{align*}Mn^{+2}\end{align*} and \begin{align*}C^{+3}\end{align*} is being oxidized to \begin{align*}C^{+4}\end{align*}.

Step 3: \begin{align*}MnO_4^- \rightarrow Mn^{2+}\end{align*}

Step 4: \begin{align*}C_2O_4^{2-} \rightarrow CO_2\end{align*}

Step 5: \begin{align*}MnO_4^- \rightarrow Mn^{2+} \quad \text{and} \quad C_2O_4^{2-} \rightarrow 2 \ CO_2\end{align*}

Step 6:

\begin{align*} MnO_4^-+ 5 \ e^- \rightarrow Mn^{2+}\\ C_2O_4^{2-} \rightarrow 2 \ CO_2+ 2 \ e^-\end{align*}

Step 7: All atoms other than \begin{align*}H\end{align*} and \begin{align*}O\end{align*} are balanced.

Step 8a: \begin{align*}MnO_4^- + 5 \ e^- + 8 \ H^+ \rightarrow Mn^{2+} + 4 \ H_2O\end{align*}

Step 8a: \begin{align*}C_2O_4^{2-} \rightarrow 2 \ CO_2 + 2 \ e^-\end{align*}

Step 9: The lowest common multiple for the electrons is 10. Therefore, we will multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.

Step 10: \begin{align*}2 \ MnO_4^- + 10 \ e^- + 16 \ H^+ \rightarrow 2 \ Mn^{2+} + 8 \ H_2O\end{align*}

Step 10: \begin{align*}5 \ C_2O_4^{2-} \rightarrow 10 \ CO_2 + 10 \ e^-\end{align*}

Step 11 and 12: \begin{align*}2 \ MnO_4^- + 16 \ H^+ + 5 \ C_2O_4^{2-} \rightarrow 2 \ Mn^{2+} + 8 \ H_2O + 10 \ CO_2\end{align*}

Example of an basic redox reaction balancing.

Given skeleton: \begin{align*}MnO_4^- + Br^- \rightarrow MnO_2 + BrO_3^- \qquad \text{(in basic solution)}\end{align*}

Step 1:

\begin{align*}\underset{{\color{red}{+7}} \ \ {\color{red}-2} \ \ }{MnO^-_4} && + && \underset{{\color{red}-1} \quad }{Br^-} && \rightarrow && \underset{{\color{red}+4} \ \ {\color{red}-2} \ \ }{MnO_2} && + && \underset{{\color{red}+5} \ \ {\color{red}-2} \quad }{BrO_3^-} && \text{(in basic solution)}\end{align*}

Step 2: \begin{align*}Mn^{+7}\end{align*} is being reduced to \begin{align*}Mn^{+4}\end{align*} and \begin{align*}Br^-\end{align*} is being oxidized to \begin{align*}Br^{+5}\end{align*}.

Step 3: \begin{align*}MnO_4^- \rightarrow MnO_2\end{align*}

Step 4: \begin{align*}Br^- \rightarrow BrO_3^-\end{align*}

Step 5: Both the atoms being oxidized and the atoms being reduced are balanced in the half-reactions.

Step 6: \begin{align*}MnO_4^- + 3 \ e^- \rightarrow MnO_2 \quad \text{and} \quad Br^- \rightarrow BrO_3^- + 6 \ e^-\end{align*}

Step 7: All atoms other than \begin{align*}H\end{align*} and \begin{align*}O\end{align*} are balanced.

Step 8b: \begin{align*}MnO_4^- + 3 \ e^- + 2 \ H_2O \rightarrow MnO_2 + 4 \ OH^-\end{align*}

Step 8b: \begin{align*}Br^- + 6 \ OH^- \rightarrow BrO_3^- + 6 \ e^- + 3 \ H_2O\end{align*}

Step 9: The LCM for the electrons is 6. Therefore, we will multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1.

Step 10: \begin{align*}2 \ MnO_4^- + 6 \ e^- + 4 \ H_2O \rightarrow 2 \ MnO_2 + 8 \ OH^-\end{align*}

Step 10: \begin{align*}Br^- + 6 \ OH^- \rightarrow BrO_3^- + 6 \ e^- + 3 \ H_2O\end{align*}

Steps 11 and 12 (Cancel electrons, \begin{align*}H_2O\end{align*}, and \begin{align*}OH^-)\end{align*}:

\begin{align*}2 \ MnO_4^- + Br^- + H_2O \rightarrow 2 \ MnO_2 + 2 \ OH^- + BrO_3^-\end{align*}

Exercises

Balance the following redox equations.

  1. \begin{align*}Br_2 + SO_2 \rightarrow Br^- + HSO_4^- \qquad \text{(in acidic solution)}\end{align*}
  2. \begin{align*}PbO_2 + Mn^{2+} \rightarrow Pb^{2+} + MnO_4^- \qquad \text{(in acidic solution)}\end{align*}
  3. \begin{align*}MnO_4^- + SO_3^{2-} \rightarrow MnO_2 + SO_4^- \qquad \text{(in basic solution)}\end{align*}
  4. \begin{align*}Zn \ + \ NO_3^- \rightarrow NH_3 + Zn(OH)_4^{2-} \qquad \text{(in basic solution)}\end{align*}
  5. \begin{align*}H_2O_2 + Cl_2O_7 \rightarrow ClO_2^- + O_2 \qquad \text{(in basic solution)}\end{align*}

Electrochemical Cells Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

Use the standard cell sketched above to answer questions 1 - 9.

1. Which electrode is the cathode?

A. \begin{align*}Pb\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

2. Which electrode is the anode?

A. \begin{align*}Pb\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

3. At which electrode will oxidation occur?

A. \begin{align*}Pb\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

4. What is the maximum voltage for this standard cell?

A. \begin{align*}0.89 \ V\end{align*}

B. \begin{align*}0.63 \ V\end{align*}

C. \begin{align*}-0.89 \ V\end{align*}

D. \begin{align*}-0.63 \ V\end{align*}

E. \begin{align*}0.50 \ V\end{align*}

5. Which way do the electrons flow in the external circuit?

A. From \begin{align*}Pb\end{align*} to \begin{align*}Zn\end{align*}.

B. From \begin{align*}Zn\end{align*} to \begin{align*}Pb\end{align*}.

C. No electron flow occurs.

6. Which way do cations flow through the salt bridge?

A. Toward the \begin{align*}Pb\end{align*} electrode.

B. Toward the \begin{align*}Zn\end{align*} electrode.

C. No cation flow occurs.

7. What happens to the cell voltage when the reaction reaches equilibrium?

A. Becomes maximum.

B. Drops to zero.

C. Becomes a positive value less than maximum.

8. Which electrode will gain mass as the cell runs?

A. \begin{align*}Pb\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

9. What happens to the cell voltage as the cell runs?

A. Remains constant.

B. Increases.

C. Decreases.

D. May increase or decrease.

Use the standard cell sketched above to answer questions 10 - 21.

10. Which electrode is the anode?

A. \begin{align*}Al\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

11. At which electrode does reduction occur?

A. \begin{align*}Al\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

12. What is the voltage of this standard cell?

A. \begin{align*}2.42 \ V\end{align*}

B. \begin{align*}-2.42 \ V\end{align*}

C. \begin{align*}0.90 \ V\end{align*}

D. \begin{align*}-0.90 \ V\end{align*}

E. \begin{align*}1.80 \ V\end{align*}

13. Which way do the electrons flow in the external circuit?

A. From \begin{align*}Al\end{align*} to \begin{align*}Zn\end{align*}.

B. From \begin{align*}Zn\end{align*} to \begin{align*}Al\end{align*}.

C. No electron flow occurs.

14. Which way do anions flows through the salt bridge?

A. Toward the \begin{align*}Al\end{align*} electrode.

B. Toward the \begin{align*}Zn\end{align*} electrode.

C. No cation flow occurs.

15. Which electrode loses mass as the cell runs?

A. \begin{align*}Al\end{align*}

B. \begin{align*}Zn\end{align*}

C. Neither.

16. How many moles of electrons pass through the external circuit in order for \begin{align*}1.00 \ mole\end{align*} of atoms to be deposited on the cathode?

A. 6

B. 3

C. 4

D. 2

E. 1

17. If 24 electrons pass through the external circuit, how many atoms of zinc must react?

A. 24

B. 12

C. 8

D. 4

E. 0

18. If 24 electrons pass through the external circuit, how many atoms of aluminum must react?

A. 24

B. 12

C. 8

D. 4

E. 0

19. What will happen to the voltage of the cell if the molarity of \begin{align*}Zn^{2+}\end{align*} is increased?

A. Increase.

B. Decrease.

C. Remain the same.

20. What will happen to the voltage of the cell if the molarity of \begin{align*}Al^{3+}\end{align*} is increased?

A. Increase.

B. Decrease.

C. Remain the same.

21. What will happen to the voltage of the cell if the salt bridge is removed?

A. Increase slightly.

B. Decrease slightly.

C. Remain the same.

D. Drop to zero.

22. In the two cells in this worksheet, there are a total of three reduction half-reaction indicated, \begin{align*}Al\end{align*}, \begin{align*}Zn\end{align*}, and \begin{align*}Pb\end{align*}. Which of these three metals is most easily oxidized?

A. \begin{align*}Al\end{align*}

B. \begin{align*}Zn\end{align*}

C. \begin{align*}Pb\end{align*}

23. Will a reaction occur if aluminum metal is placed in a solution of \begin{align*}Zn^{2+}\end{align*}?

A. Yes

B. No

24. Will a reaction occur if Pb metal is placed in a solution of \begin{align*}Al^{3+}\end{align*}?

A. Yes

B. No

25. Will a reaction occur if aluminum metal is placed in a solution of \begin{align*}Zn^{2+}\end{align*}?

A. Yes

B. No

Answers to Worksheets

  • The worksheet answer keys are available upon request. Please send an email to teachers-requests@ck12.org to request the worksheet answer keys.

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CK.SCI.ENG.TE.2.Chemistry.23.6

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