Consider the following cell reaction, Cd(s) + Hg2SO4(s) + 9/5 H2O(l) CdSO4.9/5H2O(s) + 2Hg(l). T

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

Cadmium (Cd) oxidizes from an oxidation state of 0 in the solid metal to +2 in $CdSO_4$. This means each Cd atom loses 2 electrons, so $n=2$.

Q: Why do we need to multiply $\Delta H^\circ$ by $10^3$?

The standard enthalpy change $\Delta H^\circ$ is given in kilojoules (kJ), while the electrical work term $nFE^\circ$ is calculated in joules (J). To add them, they must be in the same unit.

Q: Can we find $\Delta S^\circ$ directly from the cell potential?

Yes, if the temperature coefficient of the cell potential $(\partial E^\circ / \partial T)_P$ is known, you can use the relation $\Delta S^\circ = nF(\partial E^\circ / \partial T)_P$.

Q: What does a positive $\Delta S^\circ$ indicate here?

A positive standard entropy change indicates that the randomness or disorder of the system increases as the reaction proceeds in the forward direction.

Q: Why is the temperature taken as 298 K?

The problem states the temperature is 25°C. To use it in thermodynamic equations, it must be converted to the absolute Kelvin scale by adding 273 (25 + 273 = 298 K).

Question

Consider the following cell reaction,
$Cd(s) + Hg_2SO_4(s) + \frac{9}{5}H_2O(l) ightleftharpoons CdSO_4 \cdot \frac{9}{5}H_2O(s) + 2Hg(l)$
The value of $E_{cell}^\circ$ is $4.315	ext{ V}$ at $25^\circ	ext{C}$. If $\Delta H^\circ = -825.2	ext{ kJ mol}^{-1}$, the standard entropy change $\Delta S^\circ$ in $	ext{J K}^{-1}$ is ....... (Nearest integer)
[Given, Faraday constant = $96487	ext{ C mol}^{-1}$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We are given a fascinating electrochemical cell reaction involving Cadmium and Mercury. The problem provides us with the standard cell potential $E^\circ_{cell}$ and the standard enthalpy change $\Delta H^\circ$.

Our ultimate goal is to find the standard entropy change $\Delta S^\circ$. To do this, we need to bridge the gap between electrochemistry and classical thermodynamics.

First, let's look at the cell reaction to find the number of electrons transferred, denoted by $n$. Cadmium goes from an oxidation state of $0$ in the solid state to $+2$ in cadmium sulphate.

This means two electrons are transferred per mole of the reaction, so **$n = 2$**.

### The Master Equations
Now, we need to connect the standard cell potential with entropy and enthalpy. The master equation from thermodynamics that links these is the Gibbs free energy equation:
$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $$

We also know how Gibbs free energy relates to the standard cell potential. It is given by the fundamental electrochemical relation:
$$ \Delta G^\circ = -nFE^\circ_{cell} $$

This is our bridge! Since both expressions are equal to the standard Gibbs free energy $\Delta G^\circ$, we can equate them directly.

### Formulating the Expression
Equating the two expressions, we get:
$$ -nFE^\circ_{cell} = \Delta H^\circ - T\Delta S^\circ $$

This equation now contains all our known values and the one unknown we need to find. Let's rearrange this equation to isolate our target, the standard entropy change $\Delta S^\circ$.

Moving terms around, we get:
$$ T\Delta S^\circ = nFE^\circ_{cell} + \Delta H^\circ $$

Dividing by the absolute temperature $T$ gives us our final working formula:
$$ \Delta S^\circ = \frac{nFE^\circ_{cell} + \Delta H^\circ}{T} $$

### Final Calculation
It's time to plug in the numbers. We have $n = 2$, Faraday's constant $F = 96487 	ext{ C mol}^{-1}$, the cell potential $E^\circ_{cell} = 4.315 	ext{ V}$, and the temperature $T = 298 	ext{ K}$.

**Watch out for the units!** The enthalpy is given in kilojoules, so we must convert it to joules by multiplying by $10^3$.

Let's calculate the electrical work term $nFE^\circ_{cell}$ first:
$$ nFE^\circ_{cell} = 2 	imes 96487 	imes 4.315 \approx 832.68 	imes 10^3 	ext{ J mol}^{-1} $$

Now, let's evaluate the numerator by adding the electrical work term to the enthalpy change:
$$ 	ext{Numerator} = 832.68 	imes 10^3 + (-825.2 	imes 10^3) = 7.48 	imes 10^3 	ext{ J mol}^{-1} $$

Finally, we divide this result by the temperature $298 	ext{ K}$:
$$ \Delta S^\circ = \frac{7480}{298} \approx 25.11 	ext{ J K}^{-1} 	ext{mol}^{-1} $$

Since the question asks for the nearest integer, our final answer for the standard entropy change is **$25$**.

Enter Numerical Value:

Visualized Solution

$Number of Electrons Transferred$

$Gibbs Free Energy and Enthalpy$

$Gibbs Free Energy and Cell Potential$

$Equating the Expressions$

$Rearranging for Entropy$

$Substituting the Values$

$Calculating the Electrical Work Term$

$Evaluating the Numerator$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Electrochemical Cells

Analyzing the Setup We are given a fascinating electrochemical cell reaction involving Cadmium and Mercury

The Master Equations Now, we need to connect the standard cell potential with entropy and enthalpy

Formulating the Expression

Final Calculation It's time to plug in the numbers

Similar Questions

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, $F = 96000 C mol^{- 1}$ )

Assume a cell with the following reaction $C u (s) + 2 A g^{+} (1 \times 1 0^{- 3} M) ⟶ C u^{2 +} (0.250 M) + 2 A g (s)$ , $E_{ce l l}^{\circ} = 2.97 V$ $E_{ce l l}$ for the above reaction is ......... V. (Nearest integer) [Given : $lo g 2.5 = 0.3979, T = 298 K$ ]

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 ( $F = 96500 C mol^{- 1}$ )

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, $F = 96500 C mol^{- 1})$

Given the equilibrium constant ( $K_{c}$ ) of the reaction : $C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$ is $10 \times 1 0^{15}$ , calculate the $E_{ce l l}^{\circ}$ of this reaction at 298 K. $[2.303 \frac{R T}{F} at 298 K = 0.059 V]$

The correct option(s) about entropy ( $S$ ) is(are) [ $R = gas constant$ , $F = Faraday constant$ , $T = Temperature$ ]

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 $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Electrons Transferred

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gibbs Free Energy and Enthalpy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gibbs Free Energy and Cell Potential

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Electrical Work Term

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

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

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