Given the equilibrium constant: KC of the reaction: Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s) is 10 × 1015, calculate the E_{cell}^{0} of this reaction at 298 K [2.303 \frac{RT}{F} at 298 K = 0.059 V]

Engineering

Chemistry

EMF Calculation and Galvanic Cell

Cell Potential Free Energy and Equilibrium Constant

Nernst Equation

Question

Given the equilibrium constant:

K_C of the reaction:

Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag (s) is 10 × 10¹⁵, calculate the $E_{cell}^{0}$ of this reaction at 298 K

[2.303 $\frac{RT}{F}$ at 298 K = 0.059 V]

0.04736 mV

0.4736 mV

0.4736 V

0.04736 V

$ΔG ° = - {nFE}_{cell}^{0} = - 2 .303 RT \log K_{eq}$

$2 F E_{cell}^{o} = 2 .303 RT \log K_{eq}$

$2 E_{cell}^{o} = \frac{2 .303 RT}{F} \log 10 \times 10^{15}$

$E_{cell}^{o} = \frac{. 059}{2} \times 16 = 8 \times . 059 \Rightarrow . 472$

This question involves calculating the standard cell potential (E°_cell) from the equilibrium constant (K_C) for a redox reaction. The relationship between these two quantities is given by the Nernst equation at equilibrium.

Key Concept: At equilibrium, the cell potential (E_cell) becomes zero. The standard cell potential (E°_cell) is related to the equilibrium constant (K) by the formula:

E_{cell}^{0} = \frac{RT}{n F} \ln K

This can also be written using base-10 logarithms, which is more common for calculations:

E_{cell}^{0} = \frac{2.303 RT}{n F} \log K

Where:

R is the universal gas constant (8.314 J mol⁻¹ K⁻¹)
T is the temperature in Kelvin (298 K)
n is the number of moles of electrons transferred in the reaction
F is the Faraday constant (96485 C mol⁻¹)
K is the equilibrium constant

Step 1: Analyze the Reaction

The reaction is: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

We can break this into half-reactions to find 'n', the number of electrons transferred.

Oxidation: Cu(s) → Cu²⁺(aq) + 2e⁻

Reduction: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)

For the reaction to be balanced, 2 moles of electrons are transferred. Therefore, n = 2.

Step 2: Identify the Given Values

We are given:

K_C = 10 × 10¹⁵ = 1.0 × 10¹⁶
T = 298 K
The constant $\frac{2.303 RT}{F}$ at 298 K = 0.059 V
n = 2

Step 3: Apply the Formula

Plug the values into the formula:

E_{cell}^{0} = \frac{0.059 V}{n} \log K

E_{cell}^{0} = \frac{0.059 V}{2} \log (1.0 \times 10^{16})

Since log(10¹⁶) = 16 and log(1.0) = 0, this simplifies to:

E_{cell}^{0} = \frac{0.059 V}{2} 16

E_{cell}^{0} = 0.0295 \times 16

E_{cell}^{0} = 0.472 V

This value is closest to 0.4736 V from the options, confirming the slight difference is due to rounding the constant 0.059.

Final Answer: 0.4736 V

Detailed Solution

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Detailed Solution

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