Question

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

Short answer

For the given reactions, the balanced redox equations are: a. \(2I^{-}(aq) + MnO_{4}^{-}(aq) + 8H^+ \rightleftharpoons 2I_2(aq) + Mn^{2+}(aq) + 4H_2O\) For this reaction, \(E°_{cell} = 0.97\:V\), which indicates a spontaneous reaction under standard conditions. b. \(10F^{-}(aq) + MnO_{4}^{-}(aq) + 16H^+ \rightleftharpoons 5F_{2}(g) + Mn^{2+}(aq) + 8H_2O\) For this reaction, \(E°_{cell} = -1.36\:V\), which indicates a nonspontaneous reaction under standard conditions.

Step-by-step solution

a. Balancing the Reaction

First, we need to balance the redox reaction in aqueous (aq) solution between permanganate ion (MnO4-) and iodide ion (I-): \(MnO_{4}^{-}(aq) + I^{-}(aq) \rightleftharpoons I_{2}(aq) + Mn^{2+}(aq)\) Balance the reaction using the half-reaction method. 1. Oxidation Half-Reaction: \(I^- \rightarrow I_2\) Balance the iodine atoms: \(2I^- \rightarrow I_2\) Balance the electrons: \(2I^- \rightarrow I_2 + 2e^-\) 2. Reduction Half-Reaction: \(MnO_4^- \rightarrow Mn^{2+}\) Balance the manganese atoms: \(MnO_4^- \rightarrow Mn^{2+}\) Balance the oxygen atoms by adding water: \(MnO_4^- + 4H_2O \rightarrow Mn^{2+}\) Balance the hydrogen atoms by adding protons: \(MnO_4^- + 4H_2O + 8H^+ \rightarrow Mn^{2+}\) Balance the charge by adding electrons: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) 3. Combine the half-reactions by multiplying each half-reaction by a factor so that the number of electrons cancelled match up. Oxidation: \(2(2I^- \rightarrow I_2 + 2e^-)\) Reduction: \((5e^- + MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O)\) 4. Add the balanced half-reactions and simplify: \(2(2I^-) + MnO_4^- + 8H^+ \rightarrow 2I_2 + Mn^{2+} + 4H_2O\) The balanced redox equation is then: \(2I^{-}(aq) + MnO_{4}^{-}(aq) + 8H^+ \rightleftharpoons 2I_2(aq) + Mn^{2+}(aq) + 4H_2O\)

b. Calculating E° for Reaction a

To calculate the standard cell potential, E°, we use the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\) From Table 18.1, the standard reduction potentials (E°) are: \(E°(MnO_4^-/Mn^{2+})=1.51\:V\) \(E°(I_2/I^-) = 0.54\:V\) In this reaction, iodine is being oxidized (anode) and manganese is being reduced (cathode): \(E°_{cell} = E°(MnO_4^-/Mn^{2+}) - E°(I_2/I^-)\) \(E°_{cell} = 1.51\:V - 0.54\:V = 0.97\:V\) Since \(E°_{cell}\) is positive, the reaction is spontaneous under standard conditions.

c. Balancing the Reaction

Next, we need to balance the redox reaction in aqueous (aq) solution between permanganate ion (MnO4-) and fluoride ion (F-): \(MnO_{4}^{-}(aq) + F^{-}(aq) \rightleftharpoons F_{2}(g) + Mn^{2+}(aq)\) Follow the same steps as above to balance the reaction. 1. Oxidation Half-Reaction: \(F^- \rightarrow F_2\) Balance the fluorine atoms: \(2F^- \rightarrow F_2\) Balance the electrons: \(2F^- \rightarrow F_2 + 2e^-\) 2. Reduction Half-Reaction: \(MnO_4^- \rightarrow Mn^{2+}\) (same as in part a) \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) 3. Combine and simplify the half-reactions: \(10F^{-}(aq) + MnO_{4}^{-}(aq) + 16H^+ \rightleftharpoons 5F_2(g) + Mn^{2+}(aq) + 8H_2O\) The balanced redox equation is then: \(10F^{-}(aq) + MnO_{4}^{-}(aq) + 16H^+ \rightleftharpoons 5F_{2}(g) + Mn^{2+}(aq) + 8H_2O\)

d. Calculating E° for Reaction b

Use the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\). From Table 18.1, the standard reduction potentials (E°) are: \(E°(MnO_4^-/Mn^{2+})=1.51\rm V\) \(E°(F_2/F^{-}) = 2.87\rm V\) In this reaction, fluoride is being oxidized (anode) and manganese is being reduced (cathode): \(E°_{cell} = E°(MnO_4^-/Mn^{2+}) - E°(F_2/F^-)\) \(E°_{cell} = 1.51\:V - 2.87\:V = -1.36\:V\) Since \(E°_{cell}\) is negative, the reaction is nonspontaneous under standard conditions.

Key concepts

Standard Reduction Potential

Understanding standard reduction potentials is key in redox chemistry. It measures a substance's tendency to acquire electrons and be reduced. You find these values in tables, often set to a standard condition, like Table 18.1. Each potential is determined relative to the standard hydrogen electrode, which is assigned a potential of 0 V.

When you have a reaction, you look for the reduction potentials of the involved ions. The higher the reduction potential, the more likely the substance will gain electrons. For instance, in these exercises, manganese's reduction from permanganate has a potential of 1.51 V. This indicates a strong tendency to be reduced compared to other species like iodine's potential of 0.54 V. By comparing these values, you can predict the direction and extent of redox reactions.

Balancing Redox Equations

Balancing redox equations ensures matter and charge conservation, a fundamental aspect of chemistry. The half-reaction method is a powerful technique for achieving balance. Here’s how it typically works:

• Identify oxidation and reduction half-reactions from the overall equation.
• For each half-reaction, balance atoms of the element being oxidized or reduced. Then, balance oxygen by adding H₂O and hydrogen with H⁺, if in an acidic medium. Finally, balance the charge by adding electrons.
• Scale the half-reactions so the electrons lost and gained match, and then add them back together to form a balanced equation.

In our example, the iodide ion's oxidation results in a need to balance electrons using iodine molecules and protons for oxygen in the manganese reduction half-reaction. This painstaking balance leads to correct and meaningful equations.

Cell Potential Calculation

Cell potential provides insight into the driving force of redox reactions. It's calculated by the equation \[ E°_{cell} = E°_{cathode} - E°_{anode} \] Here, you subtract the anode's standard reduction potential from the cathode's.

In exercise (a), permanganate acts as the cathode, and iodine as the anode. You calculate the cell potential by subtracting iodine's potential (0.54 V) from permanganate's potential (1.51 V), resulting in a positive cell potential (0.97 V), signifying a spontaneous reaction.

In exercise (b), with fluoride and permanganate, the cell potential is negative (-1.36 V), indicating non-spontaneity under standard conditions. Such calculations not only tell us about reaction spontaneity but also about the feasibility and energy transfer in electrochemical cells.

Spontaneity of Reactions

The spontaneity of a redox reaction depends significantly on the sign and magnitude of its cell potential. When a reaction is spontaneous under standard conditions (\( E°_{cell} \) higher than 0), it can proceed without external influence.

A positive cell potential means the reaction releases energy, driving it forward. For example, a reaction with a cell potential of 0.97 V is spontaneous. It indicates a favorable redox process that would naturally occur.

Conversely, a reaction with a negative cell potential such as -1.36 V isn't spontaneous. It means the reaction requires an input of energy to proceed. By interpreting cell potentials, chemists can design and control electrochemical processes, influencing technology such as batteries and electroplating.