Problem 31 - Chemistry: An Atoms-Focused Approach

Question

Voltaic cells based on the following pairs of half-reactions are prepared so that all reactants and products are in their standard states. For each pair, write a balanced equation for the cell reaction, and identify which half- reaction takes place at the anode and which at the cathode. a. \(\mathrm{Hg}^{2+}(a q)+2 \mathrm{c}^{-} \rightarrow \mathrm{Hg}(\ell)\) \(\mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s)\) b. \(\mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{c}^{-} \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\) c. \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(s)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\)

Step-by-Step Solution

Answer

For each pair of half-reactions, the overall balanced cell reactions, anode, and cathode half-reactions are as follows: Pair A: Overall balanced cell reaction: \(\mathrm{Hg}^{2+}(aq)+\mathrm{Zn}(s) \rightarrow \mathrm{Hg}(\ell)+\mathrm{Zn}^{2+}(aq)\) Anode: \(\mathrm{Zn}(s)\rightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-}\) (Oxidation) Cathode: \(\mathrm{Hg}^{2+}(a q)+2 \mathrm{c}^{-} \rightarrow \mathrm{Hg}(\ell)\) (Reduction) Pair B: Overall balanced cell reaction: \(\mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{ZnO}(s) \rightarrow 2\mathrm{Ag}(s)+\mathrm{Zn}(s)+\mathrm{H}_{2}\mathrm{O}(\ell)\) Anode: \(\mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)\) (Oxidation) Cathode: \(\mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow 2\mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\) (Reduction) Pair C: Overall balanced cell reaction: \(2\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow 2\mathrm{Ni}(s)+6 \mathrm{OH}^{-}(a q)\) Anode: \(2\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightarrow 2\mathrm{Ni}(s)+4 \mathrm{OH}^{-}(a q)\) (Oxidation) Cathode: \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}(\ell)+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\) (Reduction)

Step 1: Determine the type of half-reactions

Examine each half-reaction to see if it involves oxidation or reduction: \(\mathrm{Hg}^{2+}(a q)+2 \mathrm{c}^{-} \rightarrow \mathrm{Hg}(\ell)\) → Reduction (Hg gains electrons) \(\mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s)\) → Reduction (Zn gains electrons)

Step 2: Find the Least Common Multiple (LCM) of the electron transfer

The LCM of electron transfer for both half-reactions is 2, so no need to modify the given half-reactions.

Step 3: Combine and balance the half-reactions

Combine the two reduction half-reactions and balance the charges: \[\mathrm{Hg}^{2+}(aq)+\mathrm{Zn}(s) \rightarrow \mathrm{Hg}(\ell)+\mathrm{Zn}^{2+}(aq)\]

Step 4: Identify the anode and cathode half-reactions

The oxidation half-reaction takes place at the anode. But here we have two half-reactions in reduction form. By convention, we choose the one with lower reduction potential as the anode half-reaction. So, Zn will be considered as the oxidation process. Anode Half-reaction: \(\mathrm{Zn}(s)\rightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-}\) (Oxidation) Cathode Half-reaction: \(\mathrm{Hg}^{2+}(a q)+2 \mathrm{c}^{-} \rightarrow \mathrm{Hg}(\ell)\) (Reduction) ### Pair B ###

Step 1: Determine the type of half-reactions

Examine each half-reaction to see if it involves oxidation or reduction: \(\mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)\) → Reduction (Zn gains electrons) \(\mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{H}_{2}\mathrm{O}(\ell)+2 \mathrm{c}^{-} \rightarrow 2\mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\) → Reduction (Ag gains electrons)

Step 2: Find the Least Common Multiple (LCM) of the electron transfer

The LCM of electron transfer for both half-reactions is 2, so no need to modify the given half-reactions.

Step 3: Combine and balance the half-reactions

Combine the two reduction half-reactions and balance the charges: \[\mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{ZnO}(s) \rightarrow 2\mathrm{Ag}(s)+\mathrm{Zn}(s)+\mathrm{H}_{2}\mathrm{O}(\ell)\]

Step 4: Identify the anode and cathode half-reactions

Similar to Pair A, since both half-reactions are reductions, we choose the one with lower reduction potential as the anode half-reaction (which is Zn in this case). Anode Half-reaction: \(\mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)\) (Oxidation) Cathode Half-reaction: \(\mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow 2\mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\) (Reduction) ### Pair C ###

Step 1: Determine the type of half-reactions

Examine each half-reaction to see if it involves oxidation or reduction: \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(s)+2 \mathrm{OH}^{-}(a q)\) → Reduction (Ni gains electrons) \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}(\ell)+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\) → Reduction (O gains electrons)

Step 2: Find the Least Common Multiple (LCM) of the electron transfer

The LCM of electron transfer for both half-reactions is 4, so we need to modify the first half-reaction by multiplying it by 2: \[\mathrm{2Ni}(\mathrm{OH})_{2}(s)+4 \mathrm{e}^{-} \rightarrow 2\mathrm{Ni}(s)+4 \mathrm{OH}^{-}(a q)\]

Step 3: Combine and balance the half-reactions

Combine the two reduction half-reactions and balance the charges: \[2\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow 2\mathrm{Ni}(s)+6 \mathrm{OH}^{-}(a q)\]

Step 4: Identify the anode and cathode half-reactions

Similar to Pair A and B, since both half-reactions are reductions, we choose the one with lower reduction potential as the anode half-reaction (which is Ni in this case). Anode Half-reaction: \(2\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightarrow 2\mathrm{Ni}(s)+4 \mathrm{OH}^{-}(a q)\) (Oxidation) Cathode Half-reaction: \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}(\ell)+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\) (Reduction)

Key Concepts

Half-Reactions

In electrochemistry, a half-reaction represents either the oxidation or reduction process that occurs in a voltaic cell. Each half-reaction shows either the loss or gain of electrons. For instance, in the half-reaction \( \mathrm{Hg}^{2+}(aq) + 2 \ e^- \rightarrow \mathrm{Hg}(\ell) \), mercury ions gain electrons, indicating a reduction process.
Similarly, another example is \( \mathrm{Zn}^{2+}(aq) + 2 \ e^- \rightarrow \mathrm{Zn}(s) \), where zinc ions gain electrons, showcasing yet another reduction. Understanding half-reactions helps in balancing chemical equations in cells.

Oxidation and Reduction

Oxidation and reduction are two fundamental concepts in chemistry that represent the loss and gain of electrons, respectively. Oxidation occurs when a substance loses electrons, while reduction involves a gain of electrons.
Let's consider the reaction: \( \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 \ e^- \). Here, zinc metal loses electrons, meaning it is oxidized. Conversely, in \( \mathrm{Hg}^{2+}(aq) + 2 \ e^- \rightarrow \mathrm{Hg}(\ell) \), mercury ions gain electrons, indicating reduction. These opposing processes are crucial in driving the electron flow in voltaic cells.

• Oxidation: Loss of electrons
• Reduction: Gain of electrons

Electron Transfer

Electron transfer is at the heart of how batteries work, including voltaic cells. It's the movement of electrons from one substance to another, facilitated by the difference in their reduction potentials.
This electron flow creates a current, powering devices. For example, in a reaction involving zinc and mercury, electrons move from zinc (where they are lost) to mercury (where they are gained), creating a complete circuit. Understanding electron transfer can help you grasp why current flows from the anode to the cathode in electric cells.

Anode and Cathode Identification

A key part of understanding a voltaic cell is identifying which half-reaction takes place at the anode and which at the cathode. The anode is where oxidation occurs—where electrons are lost—while the cathode is where reduction happens—where electrons are gained.
For example, in our exercise, even though both reactions showed reduction initially, we adjust by choosing the reaction with a lower reduction potential to occur at the anode (as oxidation). Here, for instance, zinc's transformation to zinc ions takes place at the anode, while mercury ions reducing to liquid mercury occurs at the cathode. Remember:

• Anode: Oxidation (electron loss)
• Cathode: Reduction (electron gain)

Properly identifying these parts is critical for predicting the direction of electron flow.