Question

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(\mathrm{X}\) and \(\mathrm{Y}\) are: $$ \begin{aligned} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow & \mathrm{Y} & &\left|E^{\circ}\right|=0.34 \mathrm{~V} \\\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow & \mathrm{X} & &\left|E^{\circ}\right|=0.25 \mathrm{~V} \end{aligned}$$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(\mathrm{X}\) to SHE. (a) Are the \(E^{\circ}\) values of the halfreactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

Short answer

(a) \(E^{\circ}\) of \(X\) is negative, \(E^{\circ}\) of \(Y\) is positive. (b) Standard emf is \(0.59\, \mathrm{V}.\)

Step-by-step solution

Interpret Electron Flow Direction

The direction of electron flow from \(X\) to \(Y\) implies that \(X\) is the anode and \(Y\) is the cathode. Electrons flow from a lower potential to a higher potential, indicating that \(Y^{2+}/Y\) has a more positive standard reduction potential than \(X^{2+}/X\).

Determine Sign of \(E^{\circ}\) for \(X\)

Since electrons flow from \(X\) to the SHE when connected, \(X\) is the anode (oxidation occurs here). This indicates \(E^{\circ}\) for \(X^{2+} + 2e^- \to X\) is negative. Therefore, \(E^{\circ}\) for \(X\) is \(-0.25\, \mathrm{V}.\)

Determine Sign of \(E^{\circ}\) for \(Y\)

With \(Y\) acting as the cathode in the \(X|Y\) cell, it must have a positive \(E^{\circ}\) value, since reduction occurs here. Thus, \(E^{\circ}\) for \(Y^{2+} + 2e^- \to Y\) is \(0.34\, \mathrm{V}.\)

Calculate the Standard EMF of the Cell

The standard EMF (\(E^{\circ}_{\text{cell}}\)) is calculated by the difference \(E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\). For the \(Y|X\) cell, \(E^{\circ}_{\text{cell}} = 0.34\, \mathrm{V} - (-0.25\, \mathrm{V}) = 0.34\, \mathrm{V} + 0.25\, \mathrm{V} = 0.59\, \mathrm{V}.\)

Key concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is a crucial concept. It refers to the tendency of a chemical species to acquire electrons and thereby be reduced. The standard reduction potential is measured under standard conditions, meaning 25 °C, 1 M concentration for each ion participating in the reaction, and a partial pressure of 1 atm for any gases involved.
The potential is given in volts, and it is denoted by the symbol \( E^{ ext{°}} \).

• A positive standard reduction potential indicates a greater tendency to gain electrons and undergo reduction.
• A negative standard reduction potential suggests that a species is more likely to lose electrons, acting as an anode in a galvanic cell.

The exercise here mentions the magnitudes of the reduction potentials for metals \( X \) and \( Y \). Understanding the signs, not just the magnitudes, provides insight into their roles as an anode or cathode when forming a cell.

Electron Flow

Electron flow is the movement of electrons from one species to another in a galvanic or voltaic cell.
In terms of electrochemistry, electrons will naturally flow from a region of lower electrochemical potential to one with a higher potential.
When the half-cells of \( X \) and \( Y \) are connected, electrons move from \( X \) to \( Y \), indicating that \( X \) has a lower standard reduction potential and thus a negative value. Consequently, \( Y \) with a higher potential value acts as a cathode where reduction occurs.

• Electron flow is essential in determining the anode and cathode of the cell.
• The direction of flow helps assign the signs for standard reduction potentials.

Understanding the direction of electron flow can enhance knowledge about the nature of reactions occurring in half-cells and dictate the overall cell potential.

Anode and Cathode

In any electrochemical cell, the anode and cathode are the two crucial electrodes.
At the anode, oxidation takes place, meaning the substance loses electrons.
Conversely, the cathode is where reduction takes place, meaning the substance gains electrons.

• In our exercise example, \( X \) serves as the anode since electrons flow from \( X \) to another entity, including the Standard Hydrogen Electrode (SHE).
• \( Y \) serves as the cathode, receiving electrons in the cell setup with \( X \).

Assigning the anode and cathode roles help us understand the flow of current and the chemical reactions occurring in each half-cell.
It is interesting to note that within a galvanic cell, the electron flow from the anode to cathode happens spontaneously due to the potential difference.

Standard EMF Calculation

The standard electromotive force (EMF) of a cell, denoted as \( E^{ ext{°}}_{\text{cell}} \), is calculated by determining the difference in potential between the cathode and the anode.
The formula is given as:\[E^{\text{°}}_{\text{cell}} = E^{\text{°}}_{\text{cathode}} - E^{\text{°}}_{\text{anode}}\]In our context, considering \( Y \) as the cathode with \( E^{\text{°}} = 0.34 \, \text{V} \) and \( X \) as the anode with \( E^{\text{°}} = -0.25 \, \text{V} \), we find:\[E^{\text{°}}_{\text{cell}} = 0.34 \, \text{V} - (-0.25 \, \text{V}) = 0.59 \, \text{V}\]

• This positive value implies that the cell reaction is spontaneous.
• Understanding the calculation of standard EMF helps to predict the feasibility and direction of electron flow in a cell.

This calculation assures that the cell operates as designed, and knowing the standard EMF gives us insights into the efficiency and potential energy output of the electrochemical cell.