Understanding Standard Electrode Potentials

In electrochemistry, a standard electrode potential ($E^\circ$) is the potential of a half-cell reaction relative to the standard hydrogen electrode (SHE) under standard conditions (298 K, 1 atm pressure, 1 M concentration). These potentials help us determine the voltage of an electrochemical cell formed by combining two half-cells.

The cell potential ($E_{\text{cell}}$) of a galvanic cell is calculated as the difference between the standard reduction potential of the cathode (where reduction occurs) and the standard reduction potential of the anode (where oxidation occurs):

\( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{cathode}) - E_{\text{red}}^\circ(\text{anode}) \)

Alternatively, it can be calculated using the standard reduction potential for the cathode and the standard oxidation potential for the anode:

\( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{cathode}) + E_{\text{ox}}^\circ(\text{anode}) \)

For a spontaneous reaction in a galvanic cell, \(E_{\text{cell}}^\circ\) must be positive. To obtain the largest possible positive \(E_{\text{cell}}^\circ\), we need to combine a half-cell with the most positive standard reduction potential (to act as the cathode) and a half-cell with the most negative standard reduction potential (which corresponds to the most positive standard oxidation potential, to act as the anode).

Analyzing the Given Half-Cells and Potentials

We are given four half-cell reactions with their associated standard potentials:

• (I) A³⁺ + A²⁺ → e^- E^o = +1.5V
• (II) B⁺ + e^- → B E^o = -0.5V
• (III) C²⁺ + e^- → C⁺ E^o = +0.5V
• (IV) D → D²⁺ + 2e^- E^o = -1.5V

Let's interpret the given potentials as standard reduction potentials, except for (IV) which is explicitly written as oxidation.

• (I) A³⁺ + e^- → A²⁺: Assuming the notation implies the standard reduction potential for the A³⁺/A²⁺ couple is +1.5V. So, \( E_{\text{red}}^\circ(\text{I}) = +1.5\text{V} \).
• (II) B⁺ + e^- → B: This is a reduction reaction. \( E_{\text{red}}^\circ(\text{II}) = -0.5\text{V} \).
• (III) C²⁺ + e^- → C⁺: This is a reduction reaction. \( E_{\text{red}}^\circ(\text{III}) = +0.5\text{V} \).
• (IV) D → D²⁺ + 2e^-: This is an oxidation reaction. The potential given is for this oxidation. \( E_{\text{ox}}^\circ(\text{IV}) = -1.5\text{V} \). The corresponding reduction potential for D²⁺ + 2e^- → D would be \( E_{\text{red}}^\circ(\text{IV}) = -E_{\text{ox}}^\circ(\text{IV}) = -(-1.5\text{V}) = +1.5\text{V} \).

Based on the standard convention where electrode potentials are listed as reduction potentials, let's assume the potentials are:

• \( E_{\text{red}}^\circ(\text{I}) = +1.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{II}) = -0.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{III}) = +0.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{IV}) = +1.5\text{V} \) (derived from \(E_{\text{ox}}^\circ(\text{IV}) = -1.5\text{V}\))

To maximize \( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{cathode}) - E_{\text{red}}^\circ(\text{anode}) \), we should choose the half-cell with the highest standard reduction potential as the cathode and the half-cell with the lowest standard reduction potential as the anode.

• Highest \( E_{\text{red}}^\circ \) is +1.5V (from I or IV).
• Lowest \( E_{\text{red}}^\circ \) is -0.5V (from II).

Combinations giving potentially largest \(E_{\text{cell}}^\circ\):

• (I) as cathode (+1.5V) and (II) as anode (-0.5V): \( E_{\text{cell}}^\circ = +1.5\text{V} - (-0.5\text{V}) = +2.0\text{V} \)
• (IV) as cathode (+1.5V) and (II) as anode (-0.5V): \( E_{\text{cell}}^\circ = +1.5\text{V} - (-0.5\text{V}) = +2.0\text{V} \)

Neither of these combinations is (I) and (IV), and they both give +2.0V.

Revisiting Interpretation based on Correct Answer

The provided correct answer is (I) and (IV). Let's see if there is an interpretation where combining (I) and (IV) yields the largest potential. If we use the derived standard reduction potentials \( E_{\text{red}}^\circ(\text{I}) = +1.5\text{V} \) and \( E_{\text{red}}^\circ(\text{IV}) = +1.5\text{V} \), combining them as cathode and anode gives \( 1.5 - 1.5 = 0\text{V} \), which is not the largest.

However, if we interpret the potential given for (IV) D → D²⁺ + 2e^- E^o = -1.5V as the standard reduction potential for that specific reaction written in the oxidation direction, this would be very unusual but leads to the correct option. Let's explore this less standard interpretation for the sake of matching the answer:

• \( E_{\text{red}}^\circ(\text{I}) = +1.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{II}) = -0.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{III}) = +0.5\text{V} \)
• \( E_{\text{red}}^\circ(\text{IV}) = -1.5\text{V} \) (Assuming the value given for oxidation is mistakenly the reduction potential)

Ordering these potentials from highest to lowest:

1. (I): +1.5V
2. (III): +0.5V
3. (II): -0.5V
4. (IV): -1.5V

To get the largest \( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{cathode}) - E_{\text{red}}^\circ(\text{anode}) \), we select the half-cell with the highest reduction potential as the cathode and the half-cell with the lowest reduction potential as the anode.

• Highest \( E_{\text{red}}^\circ \) is from (I) = +1.5V.
• Lowest \( E_{\text{red}}^\circ \) is from (IV) = -1.5V.

Combining (I) as cathode and (IV) as anode:

• Cathode (Reduction): A³⁺ + e^- → A²⁺, \( E_{\text{red}}^\circ = +1.5\text{V} \)
• Anode (Oxidation): D → D²⁺ + 2e^-. This is the reverse of the assumed reduction (IV). The potential for D → D²⁺ + 2e^- would be \( -(-1.5\text{V}) = +1.5\text{V} \) if we used oxidation potentials. If we strictly use the subtraction formula with the assumed reduction potentials: \( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{I}) - E_{\text{red}}^\circ(\text{IV}) = +1.5\text{V} - (-1.5\text{V}) = +3.0\text{V} \).

Calculating Ecell for Half-Cell Combinations

Let's calculate \(E_{\text{cell}}^\circ\) for the option (I) and (IV) using the interpretation where \(E_{\text{red}}^\circ(\text{I}) = +1.5\text{V}\) and \(E_{\text{red}}^\circ(\text{IV}) = -1.5\text{V}\).

In a galvanic cell made from (I) and (IV), (I) has the higher reduction potential (+1.5V) and acts as the cathode. (IV) has the lower reduction potential (-1.5V) and acts as the anode.

\( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{cathode}) - E_{\text{red}}^\circ(\text{anode}) \)

\( E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{I}) - E_{\text{red}}^\circ(\text{IV}) \)

\( E_{\text{cell}}^\circ = (+1.5\text{V}) - (-1.5\text{V}) \)

\( E_{\text{cell}}^\circ = +1.5\text{V} + 1.5\text{V} = +3.0\text{V} \)

Let's quickly check the Ecell for other options based on this interpretation:

• Option 1: I (+1.5V) and III (+0.5V). I is cathode, III is anode. \( E_{\text{cell}}^\circ = +1.5\text{V} - (+0.5\text{V}) = +1.0\text{V} \).
• Option 3: II (-0.5V) and IV (-1.5V). II is cathode, IV is anode. \( E_{\text{cell}}^\circ = -0.5\text{V} - (-1.5\text{V}) = -0.5\text{V} + 1.5\text{V} = +1.0\text{V} \).
• Option 4: III (+0.5V) and IV (-1.5V). III is cathode, IV is anode. \( E_{\text{cell}}^\circ = +0.5\text{V} - (-1.5\text{V}) = +0.5\text{V} + 1.5\text{V} = +2.0\text{V} \).

Comparing the potentials:

Under this specific interpretation of potential (IV), the combination of half-cells (I) and (IV) gives the largest \(E_{\text{cell}}^\circ\) value of +3.0V.

Conclusion on Largest Ecell

Assuming the standard reduction potentials are \( E_{\text{red}}^\circ(\text{I}) = +1.5\text{V} \), \( E_{\text{red}}^\circ(\text{II}) = -0.5\text{V} \), \( E_{\text{red}}^\circ(\text{III}) = +0.5\text{V} \), and \( E_{\text{red}}^\circ(\text{IV}) = -1.5\text{V} \) (deviating from the standard interpretation of oxidation potential), the combination of half-cell (I) and half-cell (IV) results in the largest standard cell potential.

Revision Table: Standard Potentials in Electrochemistry

Additional Information: Factors Affecting Ecell

The standard cell potential ($E_{\text{cell}}^\circ$) is calculated under standard conditions. In non-standard conditions, the cell potential ($E_{\text{cell}}$) is calculated using the Nernst equation:

\( E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{RT}{nF} \ln Q \)

Where:

• R is the ideal gas constant (8.314 J/mol·K)
• T is the temperature in Kelvin
• n is the number of moles of electrons transferred in the balanced overall reaction
• F is Faraday's constant (96485 C/mol)
• Q is the reaction quotient

At 298 K, the Nernst equation simplifies to:

\( E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592}{n} \log_{10} Q \)

The relative concentrations of reactants and products influence the cell potential under non-standard conditions. Increasing reactant concentration or decreasing product concentration tends to increase the cell potential for a spontaneous reaction, driving the reaction forward according to Le Chatelier's principle.