Question

For the following electrochemical cell at \(298 \mathrm{~K}\), \(\operatorname{Pt}(s)\left|\mathrm{H}_{2}(g, 1 \mathrm{bar})\right| \mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(\alpha q) \mid \mathrm{Pt}(s)\) \(E_{\text {crll }}=0.092 \mathrm{~V}\) when \(\frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^{x} .\) Given : \(E_{\mathrm{M}^{4} / \mathrm{M}^{2}}^{0}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}\) The value of \(x\) is (A) \(-2\) (B) \(-1\) (C) 1 (D) 2

Short answer

The value of \(x\) is 1.

Step-by-step solution

Write down the Nernst equation for the reaction

The Nernst equation relates the cell potential to the standard electrode potential and the ratio of the concentrations (or activities) of the substances involved in the electrochemical reaction. The Nernst equation at temperature 298 K for the given cell is: \[ E = E^0 - \frac{0.059}{n} \log\frac{[\text{M}^{2+}]}{[\text{M}^{4+}]} \]

Insert the known values into the Nernst equation

Insert the known cell potential (\(E_{\text{cell}}\)), standard electrode potential (\(E^0\)), and the relationship between the concentrations of \(\text{M}^{2+}\) and \(\text{M}^{4+}\) into the equation to find the value of \(x\). \[ 0.092 = 0.151 - \frac{0.059}{n} \log(10^x) \]

Solve for the number of electrons \(n\) transferred in the half-cell reaction

The half-cell reaction involves a change in the oxidation state from +4 to +2, which suggests a transfer of 2 electrons: \[ \text{M}^{4+} + 2e^- \rightarrow \text{M}^{2+} \] So \(n = 2\).

Solve the equation for \(x\)

Plugging in the values of \(n\) and the known constants, the equation becomes: \[ 0.092 = 0.151 - \frac{0.059}{2} \log(10^x) \] Solve this equation to find \(x\).

Isolate the logarithm and solve for \(x\)

Isolate the logarithmic part of the equation and solve for \(x\): \[ 0.059 \log(10^x) = 0.151 - 0.092 \] \[ \log(10^x) = \frac{0.059}{0.059} \] \[ \log(10^x) = 1 \] \[ 10^x = 10^1 \] \[ x = 1 \]

Key concepts

Electrochemical Cell

An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy.

Two electrodes (an anode and a cathode) and an electrolyte, which allows ions to move between the electrodes, form the basis of any electrochemical cell. The electrodes are typically made of conductive materials such as metals, and they provide the physical site for redox (reduction-oxidation) reactions to occur.

In the exercise provided, the electrochemical cell consists of a hydrogen electrode and a metal ion electrode, represented symbolically with platinum electrodes (Pt), where hydrogen (H2) gas and metal ions (M2+ and M4+) in solution are involved in the redox reactions.

Standard Electrode Potential

The standard electrode potential, denoted as E°, is a measure of the tendency of a chemical species to be reduced, and is measured under standard conditions (298 K, 1 bar pressure, and 1 M concentration).

This potential is referenced against the standard hydrogen electrode, which is assigned a potential of 0.00 V. The standard electrode potential for a reaction provides valuable information on the direction of electron flow and the likelihood of the reaction occurring.

In the textbook problem, the standard electrode potential for the metal ion couple, M4+/M2+, is given as 0.151 V. This value helps to calculate the electromotive force (EMF) of the cell and can further be used to apply the Nernst equation for finding the concentration ratio of metal ions in the exercise.

Oxidation State

Oxidation state is an indicator of the degree of oxidation (the number of electrons lost) or reduction (the number of electrons gained) of an atom in a chemical compound. In electrochemistry, the change in oxidation state during a reaction indicates how many electrons are transferred in the redox process.

For the metal ions in the provided cell, the oxidation state change from M4+ to M2+ signifies a reduction reaction, with each M4+ ion accepting two electrons to become M2+. Understanding oxidation states is crucial when using the Nernst equation, as it allows for the determination of 'n', the number of electrons involved in the electrode reaction. In the exercise, we determined that 'n' equals 2 due to the change in oxidation states, which was then used to solve for the unknown concentration ratio exponent 'x'.