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Chapter 10: Problem 116

A battery is made by connecting strips of lead and palladium and dipping each metal into a solution of its ions \(\left(\mathrm{Pb}^{2+}\right.\) and \(\mathrm{Pd}^{2+}\), respectively). Over time, the mass of the lead strip decreases and the mass of the palladium strip increases. (a) Which metal is the anode? (b) Which metal is the cathode? (c) Write the spontaneous redox reaction going on in this battery.

Short Answer

Expert verified
(a) The anode is the lead strip. (b) The cathode is the palladium strip. (c) The spontaneous redox reaction in the battery is: \( Pb(s) + Pd^{2+}(aq) \rightarrow Pb^{2+}(aq) + Pd(s) \).

Step by step solution

01

(a) Identify the Anode

Since the mass of the lead strip decreases over time, it means that lead is undergoing oxidation (losing electrons and becoming Pb²⁺ ions), so the lead is acting as the anode.
02

(b) Identify the Cathode

Similarly, since the mass of the palladium strip increases, it means that Pd²⁺ ions are getting reduced (gaining electrons) and depositing on the palladium strip. Therefore, palladium is acting as the cathode.
03

(c) Write the Spontaneous Redox Reaction

Now, let's write the half-reactions and assemble the overall spontaneous redox reaction for this battery. Oxidation half-reaction at the anode (lead): \[ Pb \rightarrow Pb^{2+} + 2 e^-\] Reduction half-reaction at the cathode (palladium): \[ Pd^{2+} + 2 e^- \rightarrow Pd\] Now, add the two half-reactions to produce the spontaneous redox reaction: \[ Pb + Pd^{2+} \rightarrow Pb^{2+} + Pd\] So, the overall spontaneous redox reaction in the battery is: \[ Pb(s) + Pd^{2+}(aq) \rightarrow Pb^{2+}(aq) + Pd(s)\]

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Most popular questions from this chapter

The reaction \(\mathrm{Cu}^{2+}+\mathrm{I}_{2} \rightarrow \mathrm{Cu}^{+}+2 \mathrm{I}^{-}\) is not possible as written. Assign an oxidation number to each atom and explain what is wrong with this reaction.

What happens to an atom's oxidation state when the atom is oxidized?

Why did chemists find it necessary to invent oxidation states?

Consider a battery made from lead, \(\mathrm{Pb}\), and copper, \(\mathrm{Cu}\), along with \(\mathrm{Pb}^{2+}\) ions and \(\mathrm{Cu}^{2+}\) ions. In such a battery, what is the oxidizing agent, and what is the reducing agent? (Hint: Start by consulting the EMF series, and determine which metal oxidizes and which metal ion gets reduced.)

The following reaction is responsible for producing electricity in your car battery (often called a lead storage battery): \(\mathrm{Pb}+\mathrm{PbO}_{2}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (a) Assign an oxidation state to each atom. (Hint: For \(\mathrm{PbSO}_{4}\), you can figure out the charge of the \(\mathrm{Pb}\) if you remember that the charge of the sulfate ion is \(-2\left(\mathrm{SO}_{4}^{2-}\right)\). Then use shortcut Rule 7 to get the oxidation state of the \(\mathrm{Pb}\) in \(\mathrm{PbSO}_{4}\).) (b) Identify the atom that gets oxidized and the atom that gets reduced. (c) Identify the reactant that is the oxidizing agent and the reactant that is the reducing agent.
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