Question

Write a balanced net ionic equation for the overall cell reaction represented by (a) \(\mathrm{Cd}\left|\mathrm{Cd}^{2+} \| \mathrm{Sb}^{3+}\right| \mathrm{Sb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cu}^{+}, \mathrm{Cu}^{2+} \| \mathrm{Mg}^{2+}\right| \mathrm{Mg}\) (c) \(\mathrm{Pt}\left|\mathrm{Cr}^{3+}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-}\right| \mathrm{Pt} \quad\) (acid)

Short answer

Question: Write the balanced net ionic equation for the overall cell reaction for the following galvanic cells: (a) \(\mathrm{Cd}\left|\mathrm{Cd}^{2+} \| \mathrm{Sb}^{3+}\right| \mathrm{Sb}\) Answer: 3Cd + 2\(\mathrm{Sb}^{3+}\) \(\rightarrow\) 3\(\mathrm{Cd}^{2+}\) + 2Sb (b) \(\mathrm{Pt}\left|\mathrm{Cu}^{+}, \mathrm{Cu}^{2+} \|\mathrm{Mg}^{2+}\right| \mathrm{Mg}\) Answer: Cu + 2\(\mathrm{Mg}^{2+}\) \(\rightarrow\) \(\mathrm{Cu}^{+}\) + 2Mg (c) Acid \(\mathrm{Pt}\left|\mathrm{Cr}^{3+}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \|\mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-}\right| \mathrm{Pt}\) Answer: Cr + 6H\(^+\) + \(\mathrm{ClO}_{3}^{-}\) \(\rightarrow\) \(\mathrm{CrO}_{4}^{2-}\) + 3H\(_{2}\)O + \(\mathrm{Cl}^{-}\)

Step-by-step solution

Identify half-reactions

At the anode (left side), the half-reaction is oxidation, where Cd will be converted to \(\mathrm{Cd}^{2+}\). At the cathode (right side), the half-reaction is reduction, where \(\mathrm{Sb}^{3+}\) will be converted to Sb.

Balance each half-reaction

Oxidation half-reaction: Cd \(\rightarrow\) \(\mathrm{Cd}^{2+}\) + 2e\(^-\) Reduction half-reaction: \(\mathrm{Sb}^{3+}\) + 3e\(^-\) \(\rightarrow\) Sb

Combine balanced half-reactions into the net ionic equation

In order to add the two balanced half-reactions together, we need to make sure they have the same number of electrons so that they cancel each other out. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 and then add them: 3Cd \(\rightarrow\) 3\(\mathrm{Cd}^{2+}\) + 6e\(^-\) 2(\(\mathrm{Sb}^{3+}\) + 3e\(^-\) \(\rightarrow\) Sb) Overall reaction: 3Cd + 6e\(^-\) + 2\(\mathrm{Sb}^{3+}\) \(\rightarrow\) 3\(\mathrm{Cd}^{2+}\) + 6e\(^-\) + 2Sb Finally, cancel out the electrons to get the balanced net ionic equation: 3Cd + 2\(\mathrm{Sb}^{3+}\) \(\rightarrow\) 3\(\mathrm{Cd}^{2+}\) + 2Sb (b) \(\mathrm{Pt}\left|\mathrm{Cu}^{+}, \mathrm{Cu}^{2+} \|\mathrm{Mg}^{2+}\right| \mathrm{Mg}\)

Identify half-reactions

At the anode (left side), there are two oxidation half-reactions: one for \(\mathrm{Cu}^{+}\)-to-\(\mathrm{Cu}^{2+}\), and the other for \(\mathrm{Cu}\)-to-\(\mathrm{Cu}^{+}\). The \(\mathrm{Cu}\)-to-\(\mathrm{Cu}^{+}\) half-reaction will occur because its standard reduction potential is more positive than the \(\mathrm{Cu}^{+}\)-to-\(\mathrm{Cu}^{2+}\). At the cathode (right side), the reduction half-reaction is where \(\mathrm{Mg}^{2+}\) will be converted to Mg.

Balance each half-reaction

Oxidation half-reaction: Cu \(\rightarrow\) \(\mathrm{Cu}^{+}\) + e\(^-\) Reduction half-reaction: \(\mathrm{Mg}^{2+}\) + 2e\(^-\) \(\rightarrow\) Mg

Combine balanced half-reactions into the net ionic equation

Multiply the reduction half-reaction by 2 and add the half-reactions: Cu \(\rightarrow\) \(\mathrm{Cu}^{+}\) + e\(^-\) 2(\(\mathrm{Mg}^{2+}\) + 2e\(^-\) \(\rightarrow\) Mg) Overall reaction: Cu + 2e\(^-\) + 2\(\mathrm{Mg}^{2+}\) \(\rightarrow\) \(\mathrm{Cu}^{+}\) + 2e\(^-\) + 2Mg Cancel out the electrons to get the balanced net ionic equation: Cu + 2\(\mathrm{Mg}^{2+}\) \(\rightarrow\) \(\mathrm{Cu}^{+}\) + 2Mg (c) Acid \(\mathrm{Pt}\left|\mathrm{Cr}^{3+}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \|\mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-}\right| \mathrm{Pt}\)

Identify half-reactions

At the anode (left side), there are two oxidation half-reactions: one for \(\mathrm{Cr}^{3+}\)-to-\((\mathrm{Cr}_{2} \mathrm{O}_{7})^{2-}\), and the other for \(\mathrm{Cr}\)-to-\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{-}\). The latter will occur as its standard reduction potential is more positive than the former. At the cathode (right side), the reduction half-reaction is where \(\mathrm{ClO}_{3}^{-}\) will be converted to Cl\(^-\) in an acidic solution.

Balance each half-reaction

Oxidation half-reaction (acidic condition): Cr \(\rightarrow\) \(\mathrm{CrO}_{4}^{2-}\) + 3H\(_{2}\)O + 6e$$^-$ Reduction half-reaction (acidic condition): 6e\(^-\) + 6H\(^+\) + \(\mathrm{Cl} \mathrm{O}_{3}^{-} \rightarrow\) 3H\(_{2}\)O + \(\mathrm{Cl}^{-}\)

Combine balanced half-reactions into the net ionic equation

Add the balanced half-reactions together: Cr + 6e\(^-\) \(\rightarrow\) \(\mathrm{CrO}_{4}^{2-}\) + 3H\(_{2}\)O + 6e$$^-$ 6e\(^-\) + 6H\(^+\) + \(\mathrm{ClO}_{3}^{-}\) \(\rightarrow\) 3H\(_{2}\)O + \(\mathrm{Cl}^{-}\) Overall reaction: Cr + 6H\(^+\) + \(\mathrm{ClO}_{3}^{-}\) \(\rightarrow\) \(\mathrm{CrO}_{4}^{2-}\) + 6H\(_{2}\)O + 6e$$^-\( + 6e\)^-\( + \)\mathrm{Cl}^{-}$ Cancel out the electrons to get the balanced net ionic equation: Cr + 6H\(^+\) + \(\mathrm{ClO}_{3}^{-}\) \(\rightarrow\) \(\mathrm{CrO}_{4}^{2-}\) + 3H\(_{2}\)O + \(\mathrm{Cl}^{-}\)