The standard electrode potential E° and its temperature coefficient (dE°/dT) for a cell are 2 V and

Q: Why is $n=2$ in this reaction?

In the given cell reaction, $\text{Zn}$ oxidizes to $\text{Zn}^{2+}$ by losing 2 electrons, and $\text{Cu}^{2+}$ reduces to $\text{Cu}$ by gaining those 2 electrons. Therefore, the number of moles of electrons transferred per mole of reaction is $n=2$.

Q: What does a negative temperature coefficient mean physically?

A negative temperature coefficient $\left(\frac{dE^{\circ}}{dT} < 0\right)$ means that the cell potential decreases as the temperature increases. Thermodynamically, this implies that the standard entropy change ($\Delta S^{\circ}$) of the cell reaction is negative.

Q: Why do we use $96000$ instead of $96500$ for Faraday's constant?

While the exact value of Faraday's constant is approximately $96485\text{ C mol}^{-1}$ (often rounded to $96500$), the problem explicitly instructs us to use $F = 96000\text{ C mol}^{-1}$ to simplify the numerical calculations.

Q: Why did we convert the final answer to kJ?

The standard units for $\Delta G$ and $\Delta H$ when calculated using Joules, Volts, and Coulombs will be in Joules per mole ($\text{J mol}^{-1}$). However, the options provided in the question are in kiloJoules per mole ($\text{kJ mol}^{-1}$), so we must divide by $1000$ to match the units.

Q: Can we use this method for any electrochemical cell?

Yes! As long as the cell operates reversibly and you know the temperature dependence of its electromotive force (EMF), you can use these exact thermodynamic relations to find $\Delta G$, $\Delta S$, and $\Delta H$ for the underlying chemical reaction.

Question

The standard electrode potential $E^{\circ}$ and its temperature coefficient $\left(\frac{dE^{\circ}}{dT}ight)$ for a cell are $2	ext{ V}$ and $-5 	imes 10^{-4}	ext{ V K}^{-1}$ at $300	ext{ K}$ respectively. The cell reaction is $	ext{Zn}(s) + 	ext{Cu}^{2+}(aq) ightarrow 	ext{Zn}^{2+}(aq) + 	ext{Cu}(s)$
The standard reaction enthalpy $(\Delta_r H^{\circ})$ at $300	ext{ K}$ in $	ext{kJ mol}^{-1}$ is,
[Use, $R = 8	ext{ J K}^{-1}	ext{ mol}^{-1}$ and $F = 96,000	ext{ C mol}^{-1}$]

Options:

(A) $-412.8$

(B) $-384.0$

(C) $206.4$

(D) $192.0$

Sigma Percentile · Accepted Answer

### The Bridge Between Two Worlds

Imagine a standard Daniell cell operating smoothly. On the left, we have a zinc anode dissolving into the solution, and on the right, a copper cathode where copper ions are depositing. The voltmeter reads a standard electrode potential of $2	ext{ V}$ at $300	ext{ K}$. But here is a fascinating catch... this voltage is not a static number; it changes with temperature!

To find the standard reaction enthalpy ($\Delta_r H^{\circ}$), we need a bridge between the electrical world of the cell and the thermal world of thermodynamics. The **Gibbs-Helmholtz equation** is our master key here. It beautifully connects Gibbs free energy, enthalpy, and entropy:

$$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$

### Decoding the Electrochemical Parameters

But how do we extract $\Delta G^{\circ}$ and $\Delta S^{\circ}$ from the cell potential? The relationships are elegantly simple. The Gibbs free energy change is directly proportional to the electrical work the cell can do:

$$\Delta G^{\circ} = -nFE^{\circ}$$

And the entropy change? It is intimately tied to how the cell's potential responds to temperature changes, known as the temperature coefficient:

$$\Delta S^{\circ} = nF \left(\frac{dE^{\circ}}{dT}ight)$$

Let us calculate $\Delta G^{\circ}$ first. In our cell reaction, zinc oxidizes to $	ext{Zn}^{2+}$, losing two electrons. So, $n = 2$. Plugging in the values of $n$, Faraday's constant ($F = 96000	ext{ C mol}^{-1}$), and the cell potential ($E^{\circ} = 2	ext{ V}$):

$$\Delta G^{\circ} = -2 	imes 96000 	imes 2 = -384000	ext{ J mol}^{-1}$$

Next, let us set up the calculation for the entropy change, $\Delta S^{\circ}$. We substitute $n = 2$, $F = 96000$, and our given temperature coefficient, which is $-5 	imes 10^{-4}	ext{ V K}^{-1}$:

$$\Delta S^{\circ} = 2 	imes 96000 	imes (-5 	imes 10^{-4}) = -96	ext{ J K}^{-1}	ext{ mol}^{-1}$$

### The Grand Finale: Calculating Enthalpy

Now, we bring back our Gibbs-Helmholtz equation. We substitute our calculated $\Delta G^{\circ}$, the temperature of $300	ext{ K}$, and our calculated $\Delta S^{\circ}$ to find the enthalpy change, $\Delta H^{\circ}$:

$$-384000 = \Delta H^{\circ} - 300 	imes (-96)$$

Negative $300$ times negative $96$ is positive $28,800$. Moving it to the other side, we subtract it from negative $384,000$:

$$\Delta H^{\circ} = -384000 - 28800 = -412800	ext{ J mol}^{-1}$$

Finally, we convert Joules to kiloJoules by dividing by one thousand. The standard reaction enthalpy is **$-412.8	ext{ kJ mol}^{-1}$**. The negative sign indicates that this is a highly exothermic reaction, releasing a significant amount of heat into its surroundings!

Select Answer:

Visualized Solution

$Visualizing the Electrochemical Cell$

$The Gibbs-Helmholtz Equation$

$Connecting Thermodynamics to Electrochemistry$

$Setting up Δ G^{\circ}$

$Calculating Δ G^{\circ}$

$Setting up Δ S^{\circ}$

$Calculating Δ S^{\circ}$

$Substituting into Gibbs-Helmholtz$

$Calculating Δ H^{\circ}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electrochemical Cells

The Bridge Between Two Worlds

Decoding the Electrochemical Parameters

The Grand Finale

Similar Questions

If the standard electrode potential for a cell is $2 V$ at $300 K$ , the equilibrium constant ( $K$ ) for the reaction, $Zn (s) + Cu^{2 +} (a q) ⇌ Zn^{2 +} (a q) + Cu (s)$ at $300 K$ is approximately ( $R = 8 JK^{- 1} mol^{- 1}$ , $F = 96000 C mol^{- 1}$ )

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction is ( $F = 96500 C mol^{- 1}$ )

For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction at $2 5^{\circ} C$ will be

The Gibbs energy change (in J) for the given reaction at $[C u^{2 +}] = [S n^{2 +}] = 1 M$ and $298 K$ is $C u (s) + S n^{2 +} (a q) ⟶ C u^{2 +} (a q) + S n (s)$ ; $(E_{S n^{2 +} / S n}^{\circ} = - 0.16 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V$ , Take, $F = 96500 C mol^{- 1})$

The cell, $Zn ∣ Zn^{2 +} (1 M) ∣∣ Cu^{2 +} (1 M) ∣ Cu$ ( $E_{cell}^{\circ} = 1.10 V$ ), was allowed to be completely discharged at 298 K. The relative concentration of $Zn^{2 +}$ to $Cu^{2 +}$ $(\frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]})$ is

Given that the standard potentials ( $E^{\circ}$ ) of $C u^{2 +} / C u$ and $C u^{+} / C u$ are $0.34 V$ and $0.522 V$ respectively, the $E^{\circ}$ of $C u^{2 +} / C u^{+}$ is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Electrochemical Cell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Gibbs-Helmholtz Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Connecting Thermodynamics to Electrochemistry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up ΔG∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔG∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up ΔS∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔS∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting into Gibbs-Helmholtz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔH∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward