Question

$${\mathrm{Pt}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Pt}(\mathrm{s})\text{ and }{\mathrm{Sn}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Sn}(\mathrm{s})$$

Short answer

First, identify the reactions happening at each electrode: Pt(s) undergoes oxidation at the anode and forms Pt²⁺(aq) + 2e⁻, while Sn²⁺(aq) undergoes reduction at the cathode and forms Sn(s). Next, calculate the standard electrode potentials (E°) for both reactions: $E{°}_{Pt}=1.18\mathrm{V}$ and $E{°}_{Sn}=-0.14\mathrm{V}$. For the anode, reverse the sign of E°, giving $E{°}_{Pt(oxidation)}=-1.18\mathrm{V}$. Finally, calculate the cell potential (E°cell) by subtracting the oxidation potential from the reduction potential: $E_{cell}^{\circ }=(-0.14)-(-1.18)=1.04\mathrm{V}$. The cell potential is +1.04 V, indicating a spontaneous reaction.

Step-by-step solution

Identify the reactions

First, let's identify the reactions happening on each electrode: At the anode (oxidation), electrons are lost, so Pt(s) must go to Pt²⁺(aq) + 2e⁻: $$\mathrm{Pt}(\mathrm{s})\to {\mathrm{Pt}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$ At the cathode (reduction), electrons are gained, so Sn²⁺(aq) + 2e⁻ must go to Sn(s): $${\mathrm{Sn}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Sn}(\mathrm{s})$$

Calculate the standard electrode potentials

Now, let's calculate the standard electrode potentials (E°) for both reactions. According to the standard reduction potentials table: $$\begin{gathered} E{°}_{Pt}=1.18\mathrm{V} \\ E{°}_{Sn}=-0.14\mathrm{V} \end{gathered}$$ Recall that for the anode (oxidation), we need to reverse the sign of E° $$E{°}_{Pt(oxidation)}=-1.18\mathrm{V}$$

Calculate the cell potential

To determine the overall cell potential (E°cell), subtract the oxidation potential from the reduction potential: $${\mathrm{E}}_{\mathrm{cell}}^{\circ }={\mathrm{E}}_{\mathrm{Sn}(reduction)}^{\circ }-{\mathrm{E}}_{\mathrm{Pt}(oxidation)}^{\circ }=(-0.14)-(-1.18)=1.04\mathrm{V}$$ The cell potential for this reaction setup is +1.04 V, which indicates that the reaction is spontaneous and will proceed as written.

Key concepts

Standard Electrode Potentials

Standard Electrode Potentials (E°) are crucial in the study of electrochemistry, as they provide a measure of the intrinsic tendency of a species to lose or gain electrons. These potentials are determined under standard conditions, which include a solute concentration of 1 M, a gas pressure of 1 atm, and a temperature of 25°C (298 K).

In the context of the given example, the standard electrode potential for platinum (Pt) is reported to be +1.18 V, indicating that Pt has a strong ability to gain electrons, or undergo reduction, in its standard state. Conversely, the potential for tin (Sn) is -0.14 V, meaning it is more likely to lose electrons, or undergo oxidation. The sign of the potential indicates whether a half-cell under standard conditions will be reduced (+) or oxidized (-) when paired with the standard hydrogen electrode.

It's important to note that when combining two half-reactions to form an overall cell, as in the provided example with platinum and tin, the electrode with the higher (more positive) electrode potential will act as the cathode and thus support the reduction process. As evident in the solution, the reversal of sign for the oxidation potential is essential since the standard potentials are originally defined for reduction.

Galvanic Cell Reactions

Galvanic cells, also known as voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. The reactions occur at two separate electrodes—the anode, where oxidation takes place, and the cathode, where reduction occurs.

In our exercise, the reaction at the anode involves platinum transitioning from solid to ion form, releasing electrons as represented by the equation $$\mathrm{Pt}(\mathrm{s})\to {\mathrm{Pt}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$, while at the cathode, tin ions gain electrons to form solid tin: $${\mathrm{Sn}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Sn}(\mathrm{s})$$.

The flow of electrons from the anode to the cathode through an external circuit generates electric current, while ions in the electrolyte balance the charge by moving between the electrodes. As the reaction proceeds, one can observe physical changes on the electrodes, such as the plating of tin and the dissolution of platinum.

Redox Equations

Redox equations succinctly represent the oxidation and reduction processes that occur in an electrochemical cell. 'Redox' is a portmanteau of reduction and oxidation, which always happen simultaneously—when one species is oxidized, another is reduced. Breaking down the overall reaction into its half-reactions helps in identifying which species is being oxidized and which is being reduced.

For instance, in the provided solution, the half-reaction $$\mathrm{Pt}(\mathrm{s})\to {\mathrm{Pt}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$ shows the oxidation process, as platinum loses electrons. Meanwhile, $${\mathrm{Sn}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Sn}(\mathrm{s})$$ illustrates the reduction process, with tin gaining electrons. Balancing these redox equations entails not only balancing the atoms but also ensuring that the charges on both sides are equal, accounting for the electrons exchanged in the process.

Both half-reactions, as well as the calculated potentials, are instrumental in determining the overall cell potential which predicts whether a redox reaction will occur spontaneously under standard conditions.