Problem 12 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the following electrochemical cell at \(298 \mathrm{~K}\), \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{bar})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M}) \| M^{4+}(\mathrm{aq}), M^{2+}(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})\) \(E_{\mathrm{ccll}}=0.092 \mathrm{~V}\) when \(\frac{\left[M^{2+}(\mathrm{aq})\right]}{\left[M^{4+}(\mathrm{aq})\right]}=10^{x} .\) [Adv. 2016] Given : \(E_{M^{4+} / M^{2+}}^{\circ}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}\) The value of \(x\) is (a) \(-2\) (b) \(-1\) (c) 1 (d) 2

Step-by-Step Solution

Answer

The value of \(x\) is 2.

Step 1: Write the Nernst Equation for the Cell

The Nernst equation for the given electrochemical cell is expressed as: \[ E_{ ext{cell}} = E^ ext{°}_{ ext{cell}} - \frac{0.059}{n} imes \log{Q} \]where \(E_{ ext{cell}} = 0.092\, \text{V}\), \(E^ ext{°}_{M^{4+}/M^{2+}} = 0.151\, \text{V}\), \(n = 2\) (because of the change in oxidation state from \(+4\) to \(+2\)), and \(Q = \frac{[M^{2+}]}{[M^{4+}]}\).

Step 2: Calculate the Standard Cell Potential

Since the hydrogen electrode is involved (with potential \(E^ ext{°}_{ ext{hydrogen}} = 0\)), and we are only given the potential for \(M^{4+}/M^{2+}\), the standard cell potential \(E^ ext{°}_{ ext{cell}}\) is:\[ E^ ext{°}_{ ext{cell}} = E^ ext{°}_{M^{4+}/M^{2+}} - E^ ext{°}_{ ext{hydrogen}} = 0.151\, \text{V} \].

Step 3: Substitute Values into Nernst Equation

Substitute known values into the Nernst Equation:\[ 0.092 = 0.151 - \frac{0.059}{2} \log\left(\frac{[M^{2+}]}{[M^{4+}]}\right) \].

Step 4: Solve for the Reaction Quotient

Rearrange the Nernst equation to solve for the logarithmic term:\[ \frac{0.059}{2} \log\left(\frac{[M^{2+}]}{[M^{4+}]}\right) = 0.151 - 0.092=0.059 \].Therefore,\[ \log\left(\frac{[M^{2+}]}{[M^{4+}]}\right) = \frac{0.059}{0.059/2} = 2 \].

Step 5: Determine the Value of x

Since \(\log\left(\frac{[M^{2+}]}{[M^{4+}]}\right) = 2\), it follows that:\[ \frac{[M^{2+}]}{[M^{4+}]} = 10^{2} \].Thus, the value of \(x\) is \(2\).

Key Concepts

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry that allows us to calculate the cell potential under non-standard conditions. The formula is given by: \[ E_{\text{cell}} = E^\text{°}_{\text{cell}} - \frac{0.059}{n} \log{Q} \] where:

• \(E_{\text{cell}}\) is the cell potential at a given moment.
• \(E^\text{°}_{\text{cell}}\) is the standard cell potential.
• \(n\) is the number of electrons transferred in the balanced reaction.
• \(Q\) is the reaction quotient, a ratio representing the concentrations of the reactants and products.

In our problem, the equation helps determine how the concentration of substances impacts the cell potential. It shows that as conditions drift from standard-state, the cell potential changes.

Standard Cell Potential

The Standard Cell Potential, denoted as \(E^\text{°}_{\text{cell}}\), is an important concept that reflects the inherent energy-driving force of an electrochemical cell under standard conditions. These conditions include all reactants and products at 1 M concentration, gases at 1 atm pressure, and the temperature at 298 K. For our exercise, the standard cell potential is the difference in electrodes' standard reduction potentials. Here, it is calculated using:\[ E^\text{°}_{\text{cell}} = E^\text{°}_{M^{4+}/M^{2+}} - E^\text{°}_{\text{hydrogen}} \]Since hydrogen's standard potential \((E^\text{°}_{\text{hydrogen}})\) is 0 V by definition, the standard cell potential becomes equivalent to \(E^\text{°}_{M^{4+}/M^{2+}}\) which is \(0.151\, V\). This represents the total potential available from the spontaneous reaction under standard conditions.

Reaction Quotient

The Reaction Quotient (\(Q\)) gives us insight into a reaction's progress and tells us how far a reaction has proceeded towards equilibrium. It is calculated similarly to the equilibrium constant (\(K\)), using the current concentrations or pressures of the reacting species:\[ Q = \frac{[\text{products}]}{[\text{reactants}]} \]For the given cell, the reaction quotient \(Q\) is expressed as:\[ Q = \frac{[M^{2+}]}{[M^{4+}]} \]During the calculation, it determines the log term in the Nernst Equation. A changing \(Q\) affects the cell potential; when \(Q\) equals the equilibrium constant (\(K\)), the reaction is at equilibrium and \(E_{\text{cell}}\) becomes zero.

Oxidation State Change

A change in the oxidation state is crucial for understanding the electron transfer in electrochemical reactions. Specifically, it reveals which species are oxidized and which are reduced during the reaction.In our electrochemical cell, the emphasis is placed on the substance changing from \(M^{4+}\) to \(M^{2+}\).

• Oxidation state change: The number of electrons transferred is defined as \(n\) in our calculations; here, \(n = 2\) electrons are involved, as it changes from a +4 to a +2 oxidation state.
• Role in Nernst Equation: This change affects the \(\frac{0.059}{n}\) term, which adjusts the impact of concentration changes on the cell potential.

Through calculations, understanding oxidation shifts contributes to correctly applying the Nernst equation in complex reactions.