day-15-electrochemistry-class12

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Concentration cell</B></primary>

<hr>

<main>

- In these cells, two like electrodes at different concentrations are dipping in the same solution. 
- Two hydrogen electrodes at unequal gas pressures immersed in the same solution of hydrogen ions constitute an electrode-concentration cell. 
- This may be represented as follows :
 - $\scriptsize{\mathrm{Pt} ; \mathrm{H}_2\left(\mathrm{P}_1\right) \mid}$ Solution of $\scriptsize{\mathrm{H}^{+}}$ions, say, $\scriptsize{\mathrm{HCl}}$ solution $\scriptsize{\mid \mathrm{H}_2\left(\mathrm{P}_2\right) ; \mathrm{Pt}}$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

- इन कोशिकाओं में, अलग-अलग सांद्रता वाले दो समान इलेक्ट्रोड एक ही घोल में डुबाए जाते हैं।
- असमान गैस दबाव पर हाइड्रोजन आयनों के एक ही घोल में डूबे दो हाइड्रोजन इलेक्ट्रोड एक इलेक्ट्रोड-सांद्रण सेल का निर्माण करते हैं।
- इसे इस प्रकार दर्शाया जा सकता है:
 - $\scriptsize{\mathrm{Pt} ; \mathrm{H}_2\left(\mathrm{P}_1\right) \mid}$ Solution of $\scriptsize{\mathrm{H}^{+}}$ions, say, $\scriptsize{\mathrm{HCl}}$ solution $\scriptsize{\mid \mathrm{H}_2\left(\mathrm{P}_2\right) ; \mathrm{Pt}}$

</div>

</main>

<hr>

<footer><span style="color:blue">Concerntration cell</span> $\rarr$ EMF calculation of concentration cell $\rarr$ Gibbs free energy and EMF $\rarr$ Enotrpy and EMf $\rarr$ EMF at equilibrium $\rarr$ EMF calculation using ΔG $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Concentration cell</B></primary>

<hr>

<main>

- Another example of the electrode-concentration cell is that of an amalgam with two different concentrations of the same metal :
$$\scriptsize{
\mathrm{Hg}-\mathrm{Pb}\left(c_1\right), \mathrm{PbSO}_4 \text { (soln.), } \mathrm{Hg}_g-\mathrm{Pb}\left(c_2\right)
}$$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

- इलेक्ट्रोड-सांद्रण सेल का एक अन्य उदाहरण एक ही धातु की अलग-अलग सांद्रता वाले मिश्रण का है:

$$\scriptsize{
\mathrm{Hg}-\mathrm{Pb}\left(c_1\right), \mathrm{PbSO}_4 \text { (soln.), } \mathrm{Hg}_g-\mathrm{Pb}\left(c_2\right)
}$$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF calculation of concentration cell</B></primary>

<hr>

<main>

The reactions occurring are :

R.H.E. : $\scriptsize{\quad 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{H}_2\left(\mathrm{p}_2\right)}$

L.H.E. :$\scriptsize{\mathrm{H}_2\left(p_1\right) \rightleftharpoons 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}}$

Overall reaction :
$$\scriptsize{
\mathrm{H}_2\left(p_1\right) \rightleftharpoons \mathrm{H}_2\left(p_2\right)
}$$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

लेकिन संतुलन पर,

होने वाली प्रतिक्रियाएँ हैं:

आर.एच.ई. : $\scriptsize{\quad 2 \mathrm{H}^{+}+2\mathrm{e}^{-}\rightleftharpoons \mathrm{H}_2\left(\mathrm{p}_2\right)}$

एल.एच.ई. :$\scriptsize{\mathrm{H}_2\left(p_1\right) \rightleftharpoons2 \mathrm{H}^{+}+2\mathrm{e}^{-}}$

समग्र प्रतिक्रिया:
$$\scriptsize{
\mathrm{H}_2\left(p_1\right) \rightleftharpoons \mathrm{H}_2\left(p_2\right)
}$$

</div>

</main>

<hr>

<footer>Concerntration cell $\rarr$ <span style="color:blue">EMF calculation of concentration cell</span> $\rarr$ Gibbs free energy and EMF $\rarr$ Enotrpy and EMf $\rarr$ EMF at equilibrium $\rarr$ EMF calculation using ΔG $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF calculation of concentration cell</B></primary>

<hr>

<main>

This reaction is evidently independent of the concentration of the electrolyte.
At moderate pressures, $\mathrm{H}_2$ can be considered to be an ideal gas so that the ratio of the activities can be considered to be equal to the ratio of the gas pressures. Hence, the Nernst equation may be written as

$$\scriptsize{
E=E^{\circ}-\frac{0.0591}{2} \log \left(p_2 / p_1\right) \text { at } 25^{\circ} \mathrm{C}
}$$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

यह प्रतिक्रिया स्पष्ट रूप से इलेक्ट्रोलाइट की सांद्रता पर निर्भर करती है।
मध्यम दबाव पर, $\mathrm{H}_2$ को एक आदर्श गैस माना जा सकता है ताकि गतिविधियों का अनुपात गैस के दबाव के अनुपात के बराबर माना जा सके। इसलिए, नर्नस्ट समीकरण को इस प्रकार लिखा जा सकता है

$$\scriptsize{
E=E^{\circ}-\frac{0.0591}{2}\log\left(p_2/p_1\right)\text{at}25^{\circ}\mathrm{C}
}$$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Gibbs free energy and EMF</B></primary>

<hr>

<main>

$$\small{ΔG = -nFE}$$
  
 - ΔG is the gibbs free enegy change
 - n is the no of electrons involved
 - F is the faraday's constant

- If the concentration of all the reacting species is unity, then $\scriptsize{E_{\text {(cell) }}=E_{\text {(cell) }}^o}$ and we have
$$\scriptsize{
\Delta_{\mathrm{r}} G^{\mathrm{o}}=-n F E_{\text {(cell) }}^{\mathrm{o}}
}$$

- $\scriptsize{\Delta_{\mathrm{r}} G^{\mathrm{o}}}$ is the standard gibbs free energy change.

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

$$\small{ΔG = -nFE}$$
  
 - ΔG गिब्स मुक्त ऊर्जा परिवर्तन है
 - n शामिल इलेक्ट्रॉनों की संख्या है
 - एफ फैराडे का स्थिरांक है

- यदि सभी प्रतिक्रियाशील प्रजातियों की सांद्रता एकता है, तो $\scriptsize{E_{\text {(सेल) }}=E_{\text {(सेल) }}^o}$ और हमारे पास है

$$\scriptsize{
\Delta_{\mathrm{r}} G^{\mathrm{o}}=-n F E_{\text {(cell) }}^{\mathrm{o}}
}$$

- $\scriptsize{\Delta_{\mathrm{r}} G^{\mathrm{o}}}$ मानक गिब्स मुक्त ऊर्जा परिवर्तन है।

</div>

</main>

<hr>

<footer>Concerntration cell $\rarr$ EMF calculation of concentration cell $\rarr$ <span style="color:blue">Gibbs free energy and EMF</span> $\rarr$ Enotrpy and EMf $\rarr$ EMF at equilibrium $\rarr$ EMF calculation using ΔG $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Enotropy and EMF</B></primary>

<hr>

<main>

$$\small{
\Delta S^{\circ}=n F\left(\partial E^{\circ} / \partial T\right)_P
}$$

###### **<ins>Derivation</ins>**

Maxwell equation:

$$\scriptsize{
\mathrm{\partial G}=\mathrm{V \partial P}-\mathrm{S\partial T}
}$$

$\scriptsize{\therefore}$ At constant pressure,

$\scriptsize{\mathrm{\partial P}=0}$

<p>
$$\scriptsize{
\begin{aligned}
& \therefore \mathrm{\partial G}=-\mathrm{S\partial T} \\
& \therefore \mathrm{S}=-\left(\frac{\mathrm{\partial G}}{\mathrm{\partial T}}\right)_{\mathrm{P}}
\end{aligned}
}$$
</p>

</div>

$$\small{
\Delta S^{\circ}=n F\left(\partial E^{\circ} / \partial T\right)_P
}$$

###### **<ins>व्युत्पत्ति</ins>**

मैक्सवेल समीकरण:

$$\scriptsize{
\mathrm{\partial G}=\mathrm{V \partial P}-\mathrm{S\partial T}
}$$

$\scriptsize{\therefore}$ निरंतर दबाव पर,

$\scriptsize{\mathrm{\partial P}=0}$

</div>

</main>

<br>

<hr>

<footer>Concerntration cell $\rarr$ EMF calculation of concentration cell $\rarr$ Gibbs free energy and EMF $\rarr$ <span style="color:blue">Enotrpy and EMf</span> $\rarr$ EMF at equilibrium $\rarr$ EMF calculation using ΔG $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Enotropy and EMF</B></primary>

<hr>

<main>

Or,

$$\scriptsize{\Delta S=-\left(\frac{\partial (\Delta G)}{\partial T}\right)_P}$$

$$\scriptsize{
\Delta S=-\left[\frac{\partial (-n E F)}{\partial T}\right]_P
}$$

<Primary>$$\scriptsize{
\therefore\left(\frac{\mathrm{\partial E}}{\mathrm{\partial T}}\right)_{\mathrm{P}}=\frac{\Delta \mathrm{S}}{\mathrm{nF}}\ \ \ \ 
}$$</Primary>

</div>

या,

$$\scriptsize{\Delta S=-\left(\frac{\partial (\Delta G)}{\partial T}\right)_P}$$

$$\scriptsize{
\Delta S=-\left[\frac{\partial (-n E F)}{\partial T}\right]_P
}$$

<Primary>$$\scriptsize{
\therefore\left(\frac{\mathrm{\partial E}}{\mathrm{\partial T}}\right)_{\mathrm{P}}=\frac{\Delta \mathrm{S}}{\mathrm{nF}}\ \ \ \ 
}$$</Primary>

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF at equilibrium</B></primary>

<hr>

<main>

Nernst equation at equilibrium can be written as

$$\scriptsize{\therefore 0 = E^o_{cell} - \frac{RT}{nF}lnK}$$

or, <Primary>$$\scriptsize{E^o_{cell} = \frac{RT}{nF}lnK}$$</Primary>

</div>

संतुलन पर नर्नस्ट समीकरण को इस प्रकार लिखा जा सकता है

$$\scriptsize{\therefore 0 = E^o_{cell} - \frac{RT}{nF}lnK}$$

या, <Primary>$$\scriptsize{E^o_{cell} = \frac{RT}{nF}lnK}$$</Primary>

</div>

</main>

<hr>

<footer>Concerntration cell $\rarr$ EMF calculation of concentration cell $\rarr$ Gibbs free energy and EMF $\rarr$ Enotrpy and EMf $\rarr$ <span style="color:blue">EMF at equilibrium</span> $\rarr$ EMF calculation using ΔG $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF calculation using ΔG</B></primary>

<hr>

<main>

Calculate the emf and $\Delta \mathrm{G}^{\circ}$ for the cell reaction at $25^{\circ} \mathrm{C}$ :

<p>
$$\scriptsize{
\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}_{(\mathrm{aq}}^{2+}\right|| \mathrm{Cd}_{\mathrm{aq}}^{2+} \mid \mathrm{Cd}_{(\mathrm{s})}
}$$
</p>

Given $\scriptsize{\mathrm{E}^{\circ} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.763}$ and $\scriptsize{\mathrm{E}^{\circ} \mathrm{Cd}^{2+} / \mathrm{Cd}=-0.403 \mathrm{V}}$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

$25^{\circ} \mathrm{C}$ पर सेल प्रतिक्रिया के लिए ईएमएफ और $\Delta \mathrm{G}^{\circ}$ की गणना करें:

<p>
$$\scriptsize{
\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}_{(\mathrm{aq}}^{2+}\right|| \mathrm{Cd}_{\mathrm{aq}}^{2+} \mid \mathrm{Cd}_{(\mathrm{s})}
}$$
</p>

Given $\scriptsize{\mathrm{E}^{\circ} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.763}$ and $\scriptsize{\mathrm{E}^{\circ} \mathrm{Cd}^{2+} / \mathrm{Cd}=-0.403 \mathrm{V}}$

</div>

</main>

<hr>

<footer>Concerntration cell $\rarr$ EMF calculation of concentration cell $\rarr$ Gibbs free energy and EMF $\rarr$ Enotrpy and EMf $\rarr$ EMF at equilibrium $\rarr$ <span style="color:blue">EMF calculation using ΔG</span> $\rarr$ Exercise</footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF calculation using ΔG</B></primary>

<hr>

<main>

<p>
$$\scriptsize{
\begin{aligned}
& \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn} ; \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.763 \mathrm{~V} \\
& \mathrm{Cd}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd} ; \mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.403 \mathrm{~V}
\end{aligned}
}$$
</p>

Cell reaction:
$$\scriptsize{
\mathrm{Cd}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Cd}+\mathrm{Zn}^{2+}
}$$

emf of the cell is, $\scriptsize{\mathrm{E}_{\text {cell }}^{\circ}=0.763-0.403=0.360 \mathrm{V}}$

$$\scriptsize{
\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}
}$$

</div>

कोशिका प्रतिक्रिया:
$$\scriptsize{
\mathrm{Cd}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Cd}+\mathrm{Zn}^{2+}
}$$

सेल का ईएमएफ है, $\scriptsize{\mathrm{E}_{\text {cell }}^{\circ}=0.763-0.403=0.360 \mathrm{V}}$

$$\scriptsize{
\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}
}$$

</p>

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>EMF calculation using ΔG</B></primary>

<hr>

<main>

where, $\scriptsize{\mathrm{n}=2}$ and $\scriptsize{\mathrm{F}=}$ faraday constant $\scriptsize{=96485 \mathrm{C} / \mathrm{mol}}$

<p>
$$\scriptsize{
\begin{aligned}
& \Delta G^{\circ}=-2 \times 96485 \times 0.36 \\
& \therefore \Delta G^{\circ}=69.5 \mathrm{~kJ}
\end{aligned}
}$$
</p>

</div>

कहाँ, $\scriptsize{\mathrm{n}=2}$ and $\scriptsize{\mathrm{F}=}$ faraday constant $\scriptsize{=96485 \mathrm{C} / \mathrm{mol}}$

<p>
$$\scriptsize{
\begin{aligned}
& \Delta G^{\circ}=-2 \times 96485 \times 0.36 \\
& \therefore \Delta G^{\circ}=69.5 \mathrm{~kJ}
\end{aligned}
}$$
</p>

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Question** 
Calculate the equilibrium constant of the reaction:

$
\begin{aligned}
Cu(s) & +2 Ag^{+}(aq) \to Cu^{2+}(aq)+2 Ag(s) \\\\\\
E   _{\text{(cell) }}^{o} & =0.46 V
\end{aligned}
$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

**सवाल** 
प्रतिक्रिया का संतुलन स्थिरांक गणना करें:

$
\begin{aligned}
Cu(s) & +2 Ag^{+}(aq) \to Cu^{2+}(aq)+2 Ag(s) \\\\\\
E   _{\text{(cell) }}^{o} & =0.46 V
\end{aligned}
$

</div>

</main>

<hr>

<footer>Concerntration cell $\rarr$ EMF calculation of concentration cell $\rarr$ Gibbs free energy and EMF $\rarr$ Enotrpy and EMf $\rarr$ EMF at equilibrium $\rarr$ EMF calculation using ΔG $\rarr$ <span style="color:blue">Exercise</span></footer>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Solution**
$E   _{\text{(cell) }}^{o}=\frac{0.059 V}{2} \log K_C=0.46 V$ or

$
\begin{aligned}
\log K_C & =\frac{0.46 V \times 2}{0.059 V}=15.6 \\\\\\
K_C & =3.92 \times 10^{15}
\end{aligned}
$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

**समाधान**
$E   _{\text{(cell) }}^{o}=\frac{0.059 V}{2} \log K_C=0.46 V$ or

$
\begin{aligned}
\log K_C & =\frac{0.46 V \times 2}{0.059 V}=15.6 \\\\\\
K_C & =3.92 \times 10^{15}
\end{aligned}
$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Solution** $\Delta_r G^{0}=-n F E   _{(\text{cell })}^{0}$

$n$ in the above equation is $2, F=96487 C mol^{-1}$ and $E   _{\text{(cell) }}^{\circ}=1.1 V$

Therefore, $\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}$

$=-21227 J mol^{-1}$

$=-212.27 kJ mol^{-1}$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

**समाधान** $\Delta_r G^{0}=-n F E   _{(\text{cell })}^{0}$

ऊपरी समीकरण में $n$ का मान $2, F=96487 C mol^{-1}$ और $E   _{\text{(cell) }}^{\circ}=1.1 V$ है

इसलिए, $\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}$

$=-21227 J mol^{-1}$

$=-212.27 kJ mol^{-1}$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Answer**

For hydrogen electrode,$H^{+}+e^{-} \longrightarrow \frac{1}{2} H_2 \text{, it is given that } pH=10$

$\therefore[H^{+}]=10^{-10} M$

Now, using Nernst equation:

$
\mathrm{H   _{(H^{+} / \frac{1}{2} H_2 )}}=E _{(H^{+} / \frac{1}{2} H_2 )}^{\ominus}-\frac{R T}{n F} \ln \frac{1}{ [H^{+} ]}$

</div>
<div  style='float: right; width:45%; text-align: left;font-size:21px'>

**उत्तर**

हाइड्रोजन इलेक्ट्रोड के लिए,$H^{+}+e^{-} \longrightarrow \frac{1}{2} H_2 \text{, यह दिया गया है कि } pH=10$

$\therefore[H^{+}]=10^{-10} M$

अब, Nernst समीकरण का उपयोग करते हुए:

$
\mathrm{H   _{(H^{+} / \frac{1}{2} H_2 )}}=E _{(H^{+} / \frac{1}{2} H_2 )}^{\ominus}-\frac{R T}{n F} \ln \frac{1}{ [H^{+} ]}$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

<p>
$
\begin{aligned}
=E   _{(H^{+} / \frac{1}{2} H_2)}^{\ominus}-\frac{0.0591}{1} \log \frac{1}{[H^{+}]} \\\\\\
=0-\frac{0.0591}{1} \log \frac{1}{[10^{-10}]} \\\\\\
=-0.0591 \log 10^{10} \\\\\\
=-0.591 V
\end{aligned}
$
</p>

</div>
<div  style='float: right; width:45%; text-align: left;font-size:18px'>

</div>

</main>

<br>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Answer**

Applying Nernst equation we have:

$ E   _{\text{(cell) }}=E _{\text{(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^{2}} $

$ =1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}} $

$ =1.05-0.02955 \log \frac{0.16}{0.000004} $

</div>

**उत्तर**

Nernst समीकरण लागू करते हैं, हमें मिलता है:

$ E   _{\text{(cell) }}=E _{\text{(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^{2}} $

$ =1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}} $

$ =1.05-0.02955 \log \frac{0.16}{0.000004} $

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Solution** $\Delta_r G^{0}=-n F E   _{(\text{cell })}^{0}$

$n$ in the above equation is $2, F=96487 C mol^{-1}$ and $E   _{\text{(cell) }}^{\circ}=1.1 V$

Therefore, $\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}$

$=-21227 J mol^{-1}$

$=-212.27 kJ mol^{-1}$

</div>

**समाधान** $\Delta_r G^{0}=-n F E   _{(\text{cell })}^{0}$

ऊपरी समीकरण में $n$ का मान $2, F=96487 C mol^{-1}$ और $E   _{\text{(cell) }}^{\circ}=1.1 V$ है

इसलिए, $\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}$

$=-21227 J mol^{-1}$

$=-212.27 kJ mol^{-1}$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Question** The cell in which the following reaction occurs:

$ 2 Fe^{3+}(aq)+2 I^{-}(aq) \to 2 Fe^{2+}(aq)+I_2(s)$ has $E   _{\text{cell }}^{o}=0.236 V$ at $298 K$.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

</div>

**सवाल** उस कोशिका में निम्नलिखित प्रतिक्रिया होती है:

$ 2 Fe^{3+}(aq)+2 I^{-}(aq) \to 2 Fe^{2+}(aq)+I_2(s)$ जिसमें $E   _{\text{cell }}^{o}=0.236 V$ होता है $298 K$ पर।

कोशिका प्रतिक्रिया की मानक गिब्स ऊर्जा और संतुलन स्थिरांक की गणना करें।

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

**Answer**

Here, $n=2, E   _{\text{cell }}^{\ominus}=0.236 V,{ _T}=298 K$

We know that:

$\Delta_r G^{\ominus}=-n \mathrm{FE_\text{cell }}^{\ominus}$

$=-2 \times 96487 \times 0.236$

$=-45541.864 J mol^{-1}$

$=-45.54 kJ mol^{-1}$

</div>

**उत्तर**

यहां, $n=2, E   _{\text{cell }}^{\ominus}=0.236 V,{ _T}=298 K$

हम जानते हैं कि:

$\Delta_r G^{\ominus}=-n \mathrm{FE_\text{cell }}^{\ominus}$

$=-2 \times 96487 \times 0.236$

$=-45541.864 J mol^{-1}$

$=-45.54 kJ mol^{-1}$

</div>

</main>

<hr>

<Success style="float:center; text-align:center;font-size:20px;"><B>Electrochemistry</B></Success>

<primary style="float:center; text-align:center;font-size:23px;"><B>Exercise</B></primary>

<hr>

<main>

Again, $\Delta_r G^{\ominus}= -2.303 R T \log K_c$

$\Rightarrow \log K_c=-\frac{\Delta_r G^{\ominus}}{2.303 R T}$

$
=-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298}
$

$=7.981$

$\therefore K_c=$ Antilog (7.981)

$=9.57 \times 10^{7}$

</div>

फिर, $\Delta_r G^{\ominus}= -2.303 R T \log K_c$

$\Rightarrow \log K_c=-\frac{\Delta_r G^{\ominus}}{2.303 R T}$

$
=-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298}
$

$=7.981$

$\therefore K_c=$ Antilog (7.981)

$=9.57 \times 10^{7}$

</div>

</main>

<hr>