Question

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow\) $$ \begin{array}{l} \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s) \\ \text { (b) } \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)+\mathrm{S}(s) \stackrel{2-}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{S}^{2-}(a q) \end{array} $$ (c) an \(\mathrm{Ag}-\mathrm{Ag}^{+}\) half-cell and an \(\mathrm{Au}-\mathrm{AuCl}_{4}^{-}\) half-cell

Short answer

Question: Calculate the standard reduction potential (E°) for the following voltaic cells: (a) MnO2(s) + 2I-(aq) ⟶ Mn2+(aq) + I2(s) + 2H2O (b) H2(g) + 2OH-(aq) + S(s) ⟶ 2H2O + S2-(aq) (c) Ag(s) + AuCl4-(aq) ⟶ Ag+(aq) + Au(s) + 4Cl-(aq) Answer: The standard cell potentials for the three voltaic cells are: (a) \(E_\text{cell}(a)^° = -0.69 \ V\) (b) \(E_\text{cell}(b)^° = 0.14 \ V\) (c) \(E_\text{cell}(c)^° = 0.20 \ V\)

Step-by-step solution

Identify the half-reactions involved in each cell

For each voltaic cell, write down the two half-reactions that take place at the anode (oxidation) and the cathode (reduction). (a) Anode: \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q) +2e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q) +2 \mathrm{H}_{2} \mathrm{O}\) Cathode: \(2 \mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) +2e^{-}\) (b) Anode: \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q) +2e^{-}\) Cathode: \(2 \mathrm{OH}^{-}(a q)+2e^{-}+\mathrm{S}(s) \longrightarrow \mathrm{S}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}\) (c) Anode: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+ e^{-}\) Cathode: \(\mathrm{AuCl}_{4}^{-}(a q) + e^{-} \longrightarrow \mathrm{Au}(s) + 4 \mathrm{Cl}^{-}(a q)\)

Find the standard reduction potentials for the half-reactions

Look up the standard reduction potentials for each half-reaction in a table of standard reduction potentials. Make sure to reverse the potential for the oxidation half-reaction when calculating the cell potential. (a) Anode: \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q) +2e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q) +2 \mathrm{H}_{2} \mathrm{O}\); \(E°_{MnO_2}=1.23 \ V\) Cathode: \(2 \mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) +2e^{-}\); \(E°_{I_2}=0.54 \ V\) (b) Anode: \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q) +2e^{-}\); \(E°_{H_2}=0.00 \ V\) Cathode: \(2 \mathrm{OH}^{-}(a q)+2e^{-}+\mathrm{S}(s) \longrightarrow \mathrm{S}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}\); \(E°_{S}=0.14 \ V\) (c) Anode: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+ e^{-}\); \(E°_{Ag}=0.80 \ V\) Cathode: \(\mathrm{AuCl}_{4}^{-}(a q) + e^{-} \longrightarrow \mathrm{Au}(s) + 4 \mathrm{Cl}^{-}(a q)\); \(E°_{AuCl_4}=1.00 \ V\)

Calculate the cell potential using the Nernst equation

The Nernst equation is given by: $$ E_\text{cell}^° = E_\text{cathode}^° - E_\text{anode}^° $$ Use this equation to calculate the cell potential for each voltaic cell in their standard state. (a) \(E_\text{cell}(a)^° = E_{I_2}^° - E_{MnO_2}^° = 0.54 \ V - 1.23 \ V = -0.69 \ V\) (b) \(E_\text{cell}(b)^° = E_{S}^° - E_{H_2}^° = 0.14 \ V - 0.00 \ V = 0.14 \ V\) (c) \(E_\text{cell}(c)^° = E_{AuCl_4}^° - E_{Ag}^° = 1.00 \ V - 0.80 \ V = 0.20 \ V\) The standard cell potentials for the three voltaic cells are: (a) \(E_\text{cell}(a)^° = -0.69 \ V\) (b) \(E_\text{cell}(b)^° = 0.14 \ V\) (c) \(E_\text{cell}(c)^° = 0.20 \ V\)

Key concepts

Voltaic Cells

Voltaic cells, also known as galvanic cells, are the foundation of batteries and are devices that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two half-cells, where oxidation occurs at the anode and reduction at the cathode. Each half-cell contains an electrode and an electrolyte and is connected by a salt bridge that allows ions to flow while maintaining electrical neutrality.

In the exercise, we're examining three different voltaic cells that involve various metals and ions. These cells are interconnected systems where electrons flow from the anode to the cathode through an external circuit, powering any electrical devices connected to it. The potential difference between the electrodes, known as the cell potential, is critical for the cell's ability to do work.

Nernst Equation

The Nernst equation allows us to calculate the cell potential at non-standard conditions (i.e., conditions where the concentrations are not 1 M for solutions or 1 atm for gases). It is a fundamental equation in electrochemistry that shows the relationship between the electrode potential of a cell, the standard electrode potential, temperature, and the activities (often approximations of concentrations) of the chemical species involved.

While the exercise provided solutions at standard conditions, the Nernst equation can be used to see how the cell potential would change if, for instance, the concentration of one of the ions changed. This equation reminds us that the power of the cell is not static — it can vary with conditions such as concentration and temperature.

Redox Reactions

Redox reactions are chemical reactions involving the transfer of electrons between two species - oxidation is the loss of electrons, while reduction is the gain of electrons. All voltaic cells operate on the principle of redox chemistry.

In our examples:

• Manganese dioxide is reduced to manganese ions while iodide ions are oxidized to iodine.
• Hydrogen gas is oxidized to hydrogen ions while sulfur is reduced to sulfide ions.
• Silver is oxidized to silver ions while the gold complex is reduced to gold.

It is the movement of electrons during these redox processes that generates the current in a voltaic cell.

Half-Reactions

Half-reactions split the overall redox reaction into two separate equations, one for oxidation and one for reduction. These half-reactions are essential for understanding and balancing redox reactions, as well as calculating the standard electrode potentials for voltaic cells.

Each half-reaction has an associated standard electrode potential, which is a measure of the tendency of a chemical species to be reduced. By convention, all standard potentials are listed for the reduction half-reaction. To find the potential for the oxidation half-reaction, as seen in the steps of the provided solution, we must reverse the sign of the standard potential. The difference in potential between the cathode and anode determines the voltage of the cell.