Example 5

The black tarnish that forms on silver objects is primarily Ag₂S. The half-reaction for reversing the tarnishing process is as follows:

$${\text{Ag}}_{\text{2}}\text{S(s)}+{\text{2e}}^{-}\to \text{2Ag(s)}+{\text{S}}^{\text{2}-}\text{(aq)}E°=-0.\text{69 V}$$

1. Referring to Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", predict which species—H₂O₂(aq), Zn(s), I⁻(aq), Sn²⁺(aq)—can reduce Ag₂S to Ag under standard conditions.
2. Of these species—H₂O₂(aq), Zn(s), I⁻(aq), Sn²⁺(aq), identify which is the strongest reducing agent in aqueous solution and thus the best candidate for a commercial product.
3. From the data in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", suggest an alternative reducing agent that is readily available, inexpensive, and possibly more effective at removing tarnish.

Given: reduction half-reaction, standard electrode potential, and list of possible reductants

Asked for: reductants for Ag₂S, strongest reductant, and potential reducing agent for removing tarnish

Strategy:

A From their positions in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", decide which species can reduce Ag₂S. Determine which species is the strongest reductant.

B Use Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to identify a reductant for Ag₂S that is a common household product.

Solution:

We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag₂S/Ag couple in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" or (2) compare E° for each species with E° for the Ag₂S/Ag couple (−0.69 V).

1. A The species in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" are arranged from top to bottom in order of increasing reducing strength. Of the four species given in the problem, I⁻(aq), Sn²⁺(aq), and H₂O₂(aq) lie above Ag₂S, and one [Zn(s)] lies below it. We can therefore conclude that Zn(s) can reduce Ag₂S(s) under standard conditions, whereas I⁻(aq), Sn²⁺(aq), and H₂O₂(aq) cannot. Sn²⁺(aq) and H₂O₂(aq) appear twice in the table: on the left side (oxidant) in one half-reaction and on the right side (reductant) in another.
2. The strongest reductant is Zn(s), the species on the right side of the half-reaction that lies closer to the bottom of Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" than the half-reactions involving I⁻(aq), Sn²⁺(aq), and H₂O₂(aq). (Commercial products that use a piece of zinc are often marketed as a “miracle product” for removing tarnish from silver. All that is required is to add warm water and salt for electrical conductivity.)
3. B Of the reductants that lie below Zn(s) in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", and therefore are stronger reductants, only one is commonly available in household products: Al(s), which is sold as aluminum foil for wrapping foods.

Exercise

Refer to Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to predict

1. which species—Sn⁴⁺(aq), Cl⁻(aq), Ag⁺(aq), Cr³⁺(aq), and/or H₂O₂(aq)—can oxidize MnO₂(s) to MNO₄⁻ under standard conditions.
2. which species—Sn⁴⁺(aq), Cl⁻(aq), Ag⁺(aq), Cr³⁺(aq), and/or H₂O₂(aq)—is the strongest oxidizing agent in aqueous solution.

Answer:

1. Ag⁺(aq); H₂O₂(aq)
2. H₂O₂(aq)

Example 6

Use the data in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to determine whether each reaction is likely to occur spontaneously under standard conditions:

1. Sn(s) + Be²⁺(aq) → Sn²⁺(aq) + Be(s)
2. MnO₂(s) + H₂O₂(aq) + 2H⁺(aq) → O₂(g) + Mn²⁺(aq) + 2H₂O(l)

Given: redox reaction and list of standard electrode potentials (Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C")

Asked for: reaction spontaneity

Strategy:

A Identify the half-reactions in each equation. Using Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", determine the standard potentials for the half-reactions in the appropriate direction.

B Use Equation 19.10 to calculate the standard cell potential for the overall reaction. From this value, determine whether the overall reaction is spontaneous.

Solution:

1. A Metallic tin is oxidized to Sn²⁺(aq), and Be²⁺(aq) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.85 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C".

   B Adding the two half-reactions gives the overall reaction:

   $$\begin{array}{lll}\text{cathode:}\hfill & {\text{MnO}}_2\text{(s)}+4{\text{H}}^{+}\text{(aq)}+2{\text{e}}^{-}\to {\text{Mn}}^{2+}\text{(aq)}+2{\text{H}}_2\text{O(l)}\hfill & E{°}_{\text{cathode}}=1.22\text{ V}\hfill \\ \text{anode:}\hfill & {\text{H}}_2{\text{O}}_2\text{(aq)}\to {\text{O}}_2\text{(g)}+2{\text{H}}^{+}\text{(aq)}+2{\text{e}}^{-}\hfill & E{°}_{\text{anode}}=0.70\text{ V}\hfill \\ \text{overall:}\hfill & {\text{MnO}}_2\text{(s)}+{\text{H}}_2{\text{O}}_2\text{(aq)}+2{\text{H}}^{+}\text{(aq)}\to \hfill & E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}\hfill \\ \hfill & {\text{O}}_2\left(\text{g}\right)+{\text{Mn}}^{2+}\text{(aq)}+2{\text{H}}_2\text{O(l)}\hfill & =-0.53\text{ V}\hfill \end{array}$$

   The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot be used to reduce Be²⁺ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn²⁺) by metallic beryllium, which has a positive value of E°_cell, will occur spontaneously.

2. A MnO₂ is the oxidant (Mn⁴⁺ is reduced to Mn²⁺), while H₂O₂ is the reductant (O²⁻ is oxidized to O₂). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C".

   B The two half-reactions and their corresponding potentials are as follows:

   $$\begin{array}{lll}\text{cathode:}\hfill & {\text{MnO}}_2\text{(s)}+4{\text{H}}^{+}\text{(aq)}+2{\text{e}}^{-}\to {\text{Mn}}^{2+}\text{(aq)}+2{\text{H}}_2\text{O(l)}\hfill & E{°}_{\text{cathode}}=1.22\text{ V}\hfill \\ \text{anode:}\hfill & {\text{H}}_2{\text{O}}_2\text{(aq)}\to {\text{O}}_2\text{(g)}+2{\text{H}}^{+}\text{(aq)}+2{\text{e}}^{-}\hfill & E{°}_{\text{anode}}=0.70\text{ V}\hfill \\ \text{overall:}\hfill & {\text{MnO}}_2\text{(s)}+{\text{H}}_2{\text{O}}_2\text{(aq)}+2{\text{H}}^{+}\text{(aq)}\to \hfill & E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}\hfill \\ \hfill & {\text{O}}_2\left(\text{g}\right)+{\text{Mn}}^{2+}\text{(aq)}+2{\text{H}}_2\text{O(l)}\hfill & =-0.53\text{ V}\hfill \end{array}$$

   The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO₂, and oxygen gas will evolve from the solution.

Exercise

Use the data in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to determine whether each reaction is likely to occur spontaneously under standard conditions:

1. 2Ce⁴⁺(aq) + 2Cl⁻(aq) → 2Ce³⁺(aq) + Cl₂(g)
2. 4MnO₂(s) + 3O₂(g) + 4OH⁻(aq) → 4MnO₄⁻(aq) + 2H₂O

Answer:

1. spontaneous (E°_cell = 0.36 V)
2. nonspontaneous (E°_cell = −0.20 V)