Problem 70 - Chemistry The Molecular Nature of Matter

Question

Determine whether the reaction: $$ \begin{aligned} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+\mathrm{Ni}(\mathrm{OH})_{2}+2 \mathrm{OH}^{-} & \longrightarrow \\ 2 \mathrm{SO}_{4}^{2-}+\mathrm{NiO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{aligned} $$ will occur spontaneously under standard state conditions. Use \(E_{\text {cell }}^{\circ}\) calculated from the data below to answer the question. $$ \begin{aligned} \mathrm{NiO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons \mathrm{Ni}(\mathrm{OH})_{2}+2 \mathrm{OH}^{-} \\ E_{\mathrm{NiO}_{2}}^{\circ}=0.49 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{8}^{2-}+2 e^{-} \rightleftharpoons 2 \mathrm{SO}_{4}^{2-} & E_{\mathrm{SO}_{4}^{2-}}^{\circ}=2.01 \mathrm{~V} \end{aligned} $$

Step-by-Step Solution

Answer

The reaction is spontaneous because the calculated cell potential is positive.

Step 1: Write Down the Half-Reactions

The reaction is composed of two half-reactions. One of the half-reactions is the reduction of NiO2 to Ni(OH)2, and the other is the reduction of S2O82- to SO42-. Write down these half-reactions separately.

Step 2: Assign Standard Reduction Potentials

Use the provided standard potentials to assign the correct values for each half-reaction. For the reduction of NiO2 to Ni(OH)2, the given standard potential is 0.49 V. For the reduction of S2O82- to SO42-, the standard potential is 2.01 V.

Step 3: Reverse the Lower Potential Half-Reaction

Since S2O82- to SO42- has a higher reduction potential, it should remain as the reduction half-reaction. The NiO2 to Ni(OH)2 half-reaction with a lower potential will be the oxidation half-reaction, so we reverse it and change the sign of its standard potential.

Step 4: Calculate the Cell Potential

Add the standard potential for the reduction half-reaction (S2O82- to SO42-) to the negated standard potential of the oxidation half-reaction (Ni(OH)2 to NiO2) to find the overall cell potential.

Step 5: Determine Spontaneity

A positive cell potential indicates a spontaneous reaction under standard state conditions. Compare the calculated cell potential to zero to determine if the reaction is spontaneous.

Key Concepts

Standard Reduction Potentials

Understanding standard reduction potentials is crucial when studying galvanic cells and spontaneity of reactions. These are values that represent the tendency of a chemical species to gain electrons and be reduced. They are measured in volts (V) and provided under standard conditions, which include a temperature of 298 K, a 1 M concentration for each ion participating in the reaction, and a pressure of 1 atmosphere for any gases involved.

Each half-reaction has its standard reduction potential, which is found in an electrochemical series. By convention, all potentials are reported relative to the standard hydrogen electrode, which is assigned a potential of 0.00 V. A higher standard reduction potential indicates a greater likelihood of the species being reduced. When comparing two half-reactions, the one with the higher standard reduction potential will naturally tend towards reduction, while the other will tend towards oxidation when connected in a galvanic cell.

Half-Reactions in Electrochemistry

Electrochemical reactions often occur in two simultaneous steps: reduction and oxidation, commonly known as redox reactions. These can be split into half-reactions that separately describe the loss of electrons (oxidation) and the gain of electrons (reduction).

In the context of the original exercise, the reduction half-reactions provide a clearer picture of the individual processes taking place at the cathode and anode. When looking at two half-reactions, it is essential to identify which one will proceed as reduction and which as oxidation. This is determined by comparing their standard reduction potentials. The half-reaction with the lower potential will be reversed to represent oxidation. When reversed, the sign of its standard reduction potential is also reversed. For clarity and simplicity, students should always write down the half-reactions and denote the anode and cathode processes distinctly.

Cell Potential Calculation

The cell potential of a galvanic cell can be determined by taking the difference between the standard reduction potential of the cathode and the anode. However, it's pivotal to remember to reverse the standard reduction potential of the anode because it is being oxidized, thus changing the sign of its potential value.

To calculate the overall cell potential, as in the given exercise, you add the standard reduction potential of the cathode half-reaction to the (reversed) standard oxidation potential (which is the negative of the standard reduction potential) of the anode half-reaction:
\[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \]
The result is the electromotive force (emf) of the galvanic cell under standard conditions. If this calculated cell potential is a positive number, the reaction is predicted to be spontaneous. Teaching this process involves ensuring students understand how to properly manipulate the sign of the potential when dealing with oxidation half-reactions, which frequently confuses beginners.