Electrochemistry
Topics to be covered
Concentration cell
इलेक्ट्रोड-सांद्रण सेल का एक अन्य उदाहरण एक ही धातु की अलग-अलग सांद्रता वाले मिश्रण का है:
Hg−Pb(c1),PbSO4 (soln.), Hgg−Pb(c2)\scriptsize{ \mathrm{Hg}-\mathrm{Pb}\left(c_1\right), \mathrm{PbSO}_4 \text { (soln.), } \mathrm{Hg}_g-\mathrm{Pb}\left(c_2\right) }Hg−Pb(c1),PbSO4 (soln.), Hgg−Pb(c2)
EMF calculation of concentration cell
The reactions occurring are :
R.H.E. : 2H++2e−⇌H2(p2)\scriptsize{\quad 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{H}_2\left(\mathrm{p}_2\right)}2H++2e−⇌H2(p2)
L.H.E. :H2(p1)⇌2H++2e−\scriptsize{\mathrm{H}_2\left(p_1\right) \rightleftharpoons 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}}H2(p1)⇌2H++2e−
Overall reaction : H2(p1)⇌H2(p2)\scriptsize{ \mathrm{H}_2\left(p_1\right) \rightleftharpoons \mathrm{H}_2\left(p_2\right) }H2(p1)⇌H2(p2)
लेकिन संतुलन पर,
होने वाली प्रतिक्रियाएँ हैं:
आर.एच.ई. : 2H++2e−⇌H2(p2)\scriptsize{\quad 2 \mathrm{H}^{+}+2\mathrm{e}^{-}\rightleftharpoons \mathrm{H}_2\left(\mathrm{p}_2\right)}2H++2e−⇌H2(p2)
एल.एच.ई. :H2(p1)⇌2H++2e−\scriptsize{\mathrm{H}_2\left(p_1\right) \rightleftharpoons2 \mathrm{H}^{+}+2\mathrm{e}^{-}}H2(p1)⇌2H++2e−
समग्र प्रतिक्रिया: H2(p1)⇌H2(p2)\scriptsize{ \mathrm{H}_2\left(p_1\right) \rightleftharpoons \mathrm{H}_2\left(p_2\right) }H2(p1)⇌H2(p2)
This reaction is evidently independent of the concentration of the electrolyte. At moderate pressures, H2\mathrm{H}_2H2 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
E=E∘−0.05912log(p2/p1) at 25∘C\scriptsize{ E=E^{\circ}-\frac{0.0591}{2} \log \left(p_2 / p_1\right) \text { at } 25^{\circ} \mathrm{C} }E=E∘−20.0591log(p2/p1) at 25∘C
यह प्रतिक्रिया स्पष्ट रूप से इलेक्ट्रोलाइट की सांद्रता पर निर्भर करती है। मध्यम दबाव पर, H2\mathrm{H}_2H2 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
E=E∘−0.05912log(p2/p1)at25∘C\scriptsize{ E=E^{\circ}-\frac{0.0591}{2}\log\left(p_2/p_1\right)\text{at}25^{\circ}\mathrm{C} }E=E∘−20.0591log(p2/p1)at25∘C
Gibbs free energy and EMF
ΔG=−nFE\small{ΔG = -nFE}ΔG=−nFE
If the concentration of all the reacting species is unity, then E(cell) =E(cell) o\scriptsize{E_{\text {(cell) }}=E_{\text {(cell) }}^o}E(cell) =E(cell) o and we have ΔrGo=−nFE(cell) o\scriptsize{ \Delta_{\mathrm{r}} G^{\mathrm{o}}=-n F E_{\text {(cell) }}^{\mathrm{o}} }ΔrGo=−nFE(cell) o
ΔrGo=−nFE(cell) o\scriptsize{ \Delta_{\mathrm{r}} G^{\mathrm{o}}=-n F E_{\text {(cell) }}^{\mathrm{o}} }ΔrGo=−nFE(cell) o
Enotropy and EMF
ΔS∘=nF(∂E∘/∂T)P\small{ \Delta S^{\circ}=n F\left(\partial E^{\circ} / \partial T\right)_P }ΔS∘=nF(∂E∘/∂T)P
Maxwell equation:
∂G=V∂P−S∂T\scriptsize{ \mathrm{\partial G}=\mathrm{V \partial P}-\mathrm{S\partial T} }∂G=V∂P−S∂T
∴\scriptsize{\therefore}∴ At constant pressure,
∂P=0\scriptsize{\mathrm{\partial P}=0}∂P=0
∴∂G=−S∂T∴S=−(∂G∂T)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} }∴∂G=−S∂T∴S=−(∂T∂G)P
मैक्सवेल समीकरण:
∴\scriptsize{\therefore}∴ निरंतर दबाव पर,
Or,
ΔS=−(∂(ΔG)∂T)P\scriptsize{\Delta S=-\left(\frac{\partial (\Delta G)}{\partial T}\right)_P}ΔS=−(∂T∂(ΔG))P
ΔS=−[∂(−nEF)∂T]P\scriptsize{ \Delta S=-\left[\frac{\partial (-n E F)}{\partial T}\right]_P }ΔS=−[∂T∂(−nEF)]P
∴(∂E∂T)P=ΔSnF \scriptsize{ \therefore\left(\frac{\mathrm{\partial E}}{\mathrm{\partial T}}\right)_{\mathrm{P}}=\frac{\Delta \mathrm{S}}{\mathrm{nF}}\ \ \ \ }∴(∂T∂E)P=nFΔS
या,
EMF at equilibrium
We know,
ΔG=−nFEcell\scriptsize{ΔG = -nFE_{cell}}ΔG=−nFEcell
At equilibrium, ΔG = 0
∴Ecell=0\scriptsize{\therefore E_{cell} = 0}∴Ecell=0
and, Q=K\scriptsize{\mathrm{Q} = \mathrm{K}}Q=K
हम जानते हैं,
ΔG=−nFEसेल\scriptsize{ΔG = -nFE_{सेल}}ΔG=−nFEसेल
संतुलन पर, ΔG = 0
और, Q=K\scriptsize{\mathrm{Q} = \mathrm{K}}Q=K
Nernst equation at equilibrium can be written as
∴0=Ecello−RTnFlnK\scriptsize{\therefore 0 = E^o_{cell} - \frac{RT}{nF}lnK}∴0=Ecello−nFRTlnK
or, Ecello=RTnFlnK\scriptsize{E^o_{cell} = \frac{RT}{nF}lnK}Ecello=nFRTlnK
संतुलन पर नर्नस्ट समीकरण को इस प्रकार लिखा जा सकता है
या, Ecello=RTnFlnK\scriptsize{E^o_{cell} = \frac{RT}{nF}lnK}Ecello=nFRTlnK
EMF calculation using ΔG
Calculate the emf and ΔG∘\Delta \mathrm{G}^{\circ}ΔG∘ for the cell reaction at 25∘C25^{\circ} \mathrm{C}25∘C :
Zn(s)∣Zn(aq2+∣∣Cdaq2+∣Cd(s)\scriptsize{ \mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}_{(\mathrm{aq}}^{2+}\right|| \mathrm{Cd}_{\mathrm{aq}}^{2+} \mid \mathrm{Cd}_{(\mathrm{s})} }Zn(s)∣Zn(aq2+∣∣Cdaq2+∣Cd(s)
Given E∘Zn2+/Zn=−0.763\scriptsize{\mathrm{E}^{\circ} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.763}E∘Zn2+/Zn=−0.763 and E∘Cd2+/Cd=−0.403V\scriptsize{\mathrm{E}^{\circ} \mathrm{Cd}^{2+} / \mathrm{Cd}=-0.403 \mathrm{V}}E∘Cd2+/Cd=−0.403V
25∘C25^{\circ} \mathrm{C}25∘C पर सेल प्रतिक्रिया के लिए ईएमएफ और ΔG∘\Delta \mathrm{G}^{\circ}ΔG∘ की गणना करें:
Zn2++2e−→Zn;EZn2+/Zn∘=−0.763 VCd2++2e−→Cd;ECd2+/Cd∘=−0.403 V\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} }Zn2++2e−→Zn;EZn2+/Zn∘=−0.763 VCd2++2e−→Cd;ECd2+/Cd∘=−0.403 V
Cell reaction: Cd2++Zn→Cd+Zn2+\scriptsize{ \mathrm{Cd}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Cd}+\mathrm{Zn}^{2+} }Cd2++Zn→Cd+Zn2+
emf of the cell is, Ecell ∘=0.763−0.403=0.360V\scriptsize{\mathrm{E}_{\text {cell }}^{\circ}=0.763-0.403=0.360 \mathrm{V}}Ecell ∘=0.763−0.403=0.360V
ΔG∘=−nFEcell ∘\scriptsize{ \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ} }ΔG∘=−nFEcell ∘
कोशिका प्रतिक्रिया: Cd2++Zn→Cd+Zn2+\scriptsize{ \mathrm{Cd}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Cd}+\mathrm{Zn}^{2+} }Cd2++Zn→Cd+Zn2+
सेल का ईएमएफ है, Ecell ∘=0.763−0.403=0.360V\scriptsize{\mathrm{E}_{\text {cell }}^{\circ}=0.763-0.403=0.360 \mathrm{V}}Ecell ∘=0.763−0.403=0.360V
where, n=2\scriptsize{\mathrm{n}=2}n=2 and F=\scriptsize{\mathrm{F}=}F= faraday constant =96485C/mol\scriptsize{=96485 \mathrm{C} / \mathrm{mol}}=96485C/mol
ΔG∘=−2×96485×0.36∴ΔG∘=69.5 kJ\scriptsize{ \begin{aligned} & \Delta G^{\circ}=-2 \times 96485 \times 0.36 \\ & \therefore \Delta G^{\circ}=69.5 \mathrm{~kJ} \end{aligned} }ΔG∘=−2×96485×0.36∴ΔG∘=69.5 kJ
कहाँ, n=2\scriptsize{\mathrm{n}=2}n=2 and F=\scriptsize{\mathrm{F}=}F= faraday constant =96485C/mol\scriptsize{=96485 \mathrm{C} / \mathrm{mol}}=96485C/mol
Exercise
Question Calculate the equilibrium constant of the reaction:
Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) E(cell) o=0.46V \begin{aligned} Cu(s) & +2 Ag^{+}(aq) \to Cu^{2+}(aq)+2 Ag(s) \\\ E _{\text{(cell) }}^{o} & =0.46 V \end{aligned} Cu(s) E(cell) o+2Ag+(aq)→Cu2+(aq)+2Ag(s)=0.46V
सवाल प्रतिक्रिया का संतुलन स्थिरांक गणना करें:
Solution E(cell) o=0.059V2logKC=0.46VE _{\text{(cell) }}^{o}=\frac{0.059 V}{2} \log K_C=0.46 VE(cell) o=20.059VlogKC=0.46V or
logKC=0.46V×20.059V=15.6 KC=3.92×1015 \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} logKC KC=0.059V0.46V×2=15.6=3.92×1015
समाधान E(cell) o=0.059V2logKC=0.46VE _{\text{(cell) }}^{o}=\frac{0.059 V}{2} \log K_C=0.46 VE(cell) o=20.059VlogKC=0.46V or
Solution The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction:
Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s) Zn(s)+Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq)+Cu(s) Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s)
सवाल Daniell कोशिका के लिए मानक इलेक्ट्रोड पोटेंशियल 1.1V है। प्रतिक्रिया के लिए मानक गिब्स ऊर्जा की गणना करें:
Solution ΔrG0=−nFE(cell )0\Delta_r G^{0}=-n F E _{(\text{cell })}^{0}ΔrG0=−nFE(cell )0
nnn in the above equation is 2,F=96487Cmol−12, F=96487 C mol^{-1}2,F=96487Cmol−1 and E(cell) ∘=1.1VE _{\text{(cell) }}^{\circ}=1.1 VE(cell) ∘=1.1V
Therefore, ΔrG0=−2×1.1V×96487Cmol−1\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}ΔrG0=−2×1.1V×96487Cmol−1
=−21227Jmol−1=-21227 J mol^{-1}=−21227Jmol−1
=−212.27kJmol−1=-212.27 kJ mol^{-1}=−212.27kJmol−1
समाधान ΔrG0=−nFE(cell )0\Delta_r G^{0}=-n F E _{(\text{cell })}^{0}ΔrG0=−nFE(cell )0
ऊपरी समीकरण में nnn का मान 2,F=96487Cmol−12, F=96487 C mol^{-1}2,F=96487Cmol−1 और E(cell) ∘=1.1VE _{\text{(cell) }}^{\circ}=1.1 VE(cell) ∘=1.1V है
इसलिए, ΔrG0=−2×1.1V×96487Cmol−1\Delta_r G^{0}=-2 \times 1.1 V \times 96487 C mol^{-1}ΔrG0=−2×1.1V×96487Cmol−1
Question Calculate the potential of hydrogen electrode in contact with a solution whose pHpHpH is 10 .
सवाल हाइड्रोजन इलेक्ट्रोड की संभाव्यता की गणना करें जो एक समाधान के संपर्क में है जिसका pHpHpH 10 है।
Answer
For hydrogen electrode,H++e−⟶12H2, it is given that pH=10H^{+}+e^{-} \longrightarrow \frac{1}{2} H_2 \text{, it is given that } pH=10H++e−⟶21H2, it is given that pH=10
∴[H+]=10−10M\therefore[H^{+}]=10^{-10} M∴[H+]=10−10M
Now, using Nernst equation:
H(H+/12H2)=E(H+/12H2)⊖−RTnFln1[H+] \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^{+} ]}H(H+/21H2)=E(H+/21H2)⊖−nFRTln[H+]1
उत्तर
हाइड्रोजन इलेक्ट्रोड के लिए,H++e−⟶12H2, यह दिया गया है कि pH=10H^{+}+e^{-} \longrightarrow \frac{1}{2} H_2 \text{, यह दिया गया है कि } pH=10H++e−⟶21H2, यह दिया गया है कि pH=10
अब, Nernst समीकरण का उपयोग करते हुए:
=E(H+/12H2)⊖−0.05911log1[H+]=0−0.05911log1[10−10]=−0.0591log1010=−0.591V \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} =E(H+/21H2)⊖−10.0591log[H+]1=0−10.0591log[10−10]1=−0.0591log1010=−0.591V
Question Calculate the emf of the cell in which the following reaction takes place:
Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s)Ni(s)+2 Ag^{+}(0.002 M) \to Ni^{2+}(0.160 M)+2 Ag(s)Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s)
Given that Ecell o=1.05VE _{\text{cell }}^{o}=1.05 VEcell o=1.05V
सवाल उस कोशिका की emf की गणना करें जिसमें निम्नलिखित प्रतिक्रिया होती है:
यह दिया गया है कि Ecell o=1.05VE _{\text{cell }}^{o}=1.05 VEcell o=1.05V
Applying Nernst equation we have:
E(cell) =E(cell) ⊖−0.0591nlog[Ni2+][Ag+]2 E _{\text{(cell) }}=E _{\text{(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^{2}} E(cell) =E(cell) ⊖−n0.0591log[Ag+]2[Ni2+]
=1.05−0.05912log(0.160)(0.002)2 =1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}} =1.05−20.0591log(0.002)2(0.160)
=1.05−0.02955log0.160.000004 =1.05-0.02955 \log \frac{0.16}{0.000004} =1.05−0.02955log0.0000040.16
Nernst समीकरण लागू करते हैं, हमें मिलता है:
=1.05−0.02955log4×104 =1.05-0.02955 \log 4 \times 10^{4} =1.05−0.02955log4×104
=1.05−0.02955(log10000+log4) =1.05-0.02955(\log 10000+\log 4) =1.05−0.02955(log10000+log4)
=1.05−0.02955(4+0.6021) =1.05-0.02955(4+0.6021) =1.05−0.02955(4+0.6021)
=0.914 V
Question The cell in which the following reaction occurs:
2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(s) 2 Fe^{3+}(aq)+2 I^{-}(aq) \to 2 Fe^{2+}(aq)+I_2(s)2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(s) has Ecell o=0.236VE _{\text{cell }}^{o}=0.236 VEcell o=0.236V at 298K298 K298K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
सवाल उस कोशिका में निम्नलिखित प्रतिक्रिया होती है:
2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(s) 2 Fe^{3+}(aq)+2 I^{-}(aq) \to 2 Fe^{2+}(aq)+I_2(s)2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(s) जिसमें Ecell o=0.236VE _{\text{cell }}^{o}=0.236 VEcell o=0.236V होता है 298K298 K298K पर।
कोशिका प्रतिक्रिया की मानक गिब्स ऊर्जा और संतुलन स्थिरांक की गणना करें।
Here, n=2,Ecell ⊖=0.236V,T=298Kn=2, E _{\text{cell }}^{\ominus}=0.236 V,{ _T}=298 Kn=2,Ecell ⊖=0.236V,T=298K
We know that:
ΔrG⊖=−nFEcell ⊖\Delta_r G^{\ominus}=-n \mathrm{FE_\text{cell }}^{\ominus}ΔrG⊖=−nFEcell ⊖
=−2×96487×0.236=-2 \times 96487 \times 0.236=−2×96487×0.236
=−45541.864Jmol−1=-45541.864 J mol^{-1}=−45541.864Jmol−1
=−45.54kJmol−1=-45.54 kJ mol^{-1}=−45.54kJmol−1
यहां, n=2,Ecell ⊖=0.236V,T=298Kn=2, E _{\text{cell }}^{\ominus}=0.236 V,{ _T}=298 Kn=2,Ecell ⊖=0.236V,T=298K
हम जानते हैं कि:
Again, ΔrG⊖=−2.303RTlogKc\Delta_r G^{\ominus}= -2.303 R T \log K_cΔrG⊖=−2.303RTlogKc
⇒logKc=−ΔrG⊖2.303RT\Rightarrow \log K_c=-\frac{\Delta_r G^{\ominus}}{2.303 R T}⇒logKc=−2.303RTΔrG⊖
=−−45.54×1032.303×8.314×298 =-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298} =−2.303×8.314×298−45.54×103
=7.981=7.981=7.981
∴Kc=\therefore K_c=∴Kc= Antilog (7.981)
=9.57×107=9.57 \times 10^{7}=9.57×107
फिर, ΔrG⊖=−2.303RTlogKc\Delta_r G^{\ominus}= -2.303 R T \log K_cΔrG⊖=−2.303RTlogKc