The equilibrium constant of the reaction;\$Cu\left( s \right) +2{ Ag }^{ + }\left( aq \right) \longrightarrow { Cu }^{ 2+ }\left( aq \right) +2Ag\left( s \right) \$\${ E }^{ o }=0.46V\$ at \$298 K\$

\$Cu\left( s \right) +2{ Ag }^{ 2+ }\left( aq \right) \longrightarrow { Cu }^{ 2+ }\left( aq \right) +2Ag\left( s \right)\$\$ { E }^{ o }=0.46 V\$ at \$298 K\$\$\log { { K }_{ C }= } \dfrac { nF{ E }^{ o } }{ RT } \$ \$=\dfrac { 2\times 0.46 }{ 0.059 }\$\$ { K }_{ C }=4\times { 10 }^{ 15 }\$

Write the Nernst equation and emf of the following cells at \$298\ K\$:(i) \$Mg(s) | Mg^{+2}(0.001\ M)\parallel Cu^{+2}(0.0001\ M)|Cu(s)\$.(ii) \$Fe(s) | Fe^{+2} (0.001\ M) \parallel H^{+}(1M)|H_{2}(g) (1bar) | Pt(s)\$(iii) \$Sn(s) | Sn^{2+} (0.050\ M)\parallel H^{+}(0.020\ M)|H_{2}(g) (1 bar)| Pt(s)\$(iv) \$Pt(s)| Br_{2}(l)| Br^{-}(0.010\ M)\parallel H^{+}(0.030\ M)| H_{2}(g) (1 bar)| Pt(s)\$.

\$(i)\$ \$\displaystyle E^o = 0.34 - (-2.37) = 2.71\$ V\$\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Mg^{2+}]}{ [Cu^{2+}]}\$\$\displaystyle E = 2.71 -\dfrac {0.059}{2} log \dfrac {0.001}{(0.0001)}\$\$\displaystyle E = 2.6805\$ V\$(ii)\$ \$\displaystyle E^o = 0-(-0.44) = 0.44\$ V\$\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Fe^{2+}]}{ [H^+]^2}\$\$\displaystyle E = 0.44 -\dfrac {0.059}{2} log \dfrac {0.001}{(1)^2} \$\$\displaystyle E = 0.5285\$ V\$(iii)\$ \$\displaystyle E^o = 0-(-0.14) = 0.14\$ V\$\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Sn^{2+}]} { [H^+]^2}\$\$\displaystyle E = 0.14 -\dfrac {0.059}{2} log \dfrac {0.050}{(0.020)^2} \$\$\displaystyle E = 0.08\$ V\$(iv)\$ \$\displaystyle E^o = 0-1.08 =- 1.08\$ V\$\displaystyle E = E^0 - \dfrac {0.059}{2} log \dfrac {1}{[Br^-]^2 [H^+]^2}\$\$\displaystyle E = -1.08 -\dfrac {0.059}{2} log \dfrac {1}{(0.01)^2 (0.030)^2}\$\$\displaystyle E = -1.288\$ V

Represent the cell in which the following reaction takes place \$Mg\left( s \right) + 2A{g^ + }\left( {0.0001M} \right) \to M{g^{2 + }}\left( {0.130M} \right) + 2Ag\left( s \right)\$Calculate its \${E_{\left( {cell} \right)}}\,{\text{if}}\,E_{\left( {cell} \right)}^o = 3.17V\$

\$Mg+2Ag+(0.0001M)\longrightarrow Mg^{2+}(0.130M)+2Ag\$\$E_{cell}=E^o-\cfrac{0.0591}{2}\log\left[\cfrac{(Mg^{2+})}{(Ag^+)^2}\right]\$\$E_{cell}=8.17-\cfrac{0.0591}{2}\log\left[\cfrac{0.130}{10^{-8}}\right]=2.959\$ \$V\$

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:(i) \$2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd\$(ii) \$Fe^{2+}(aq) + Ag^{+} (aq) \rightarrow Fe^{3+}(aq) + Ag(s)\$Calculate the \$\Delta_{r}G^{\ominus}\$ and equilibrium constant of the reactions

(i) \$\displaystyle E^o_{cell} = E^0_{cathode} - E^0_{anode} =-0.40- (-0.74) = 0.34\$ V\$\displaystyle \Delta G^0 = -nFE^o_{cell} = - 6 \times 96500 \times 0.34 = 196860 \: J/mol = 196.86 \: kJ/mol \$\$\displaystyle log\ K_c = \dfrac {nE^o_{cell}}{0.059} = \dfrac {6 \times 0.34}{0.059} = 34.576\$\$\displaystyle K_c = 3.76 \times 10^{34}\$(ii) \$\displaystyle E^o_{cell} = E^0_{cathode} - E^0_{anode} =0.80- (-0.77) = 0.03\$ V\$\displaystyle \Delta G^0 = -nFE^o_{cell} = - 1 \times 96500 \times 0.03 = 2895 \: J/mol = 2.895 \: kJ/mol \$\$\displaystyle log\ K_c = \dfrac {nE^o_{cell}}{0.059} = \dfrac {1 \times 0.03}{0.059} = 0.508\$\$\displaystyle K_c = 3.22\$

The Gibbs energy for the decomposition of \$Al_2O_3\$ at \$500\$ \$^oC\$ is as follows:\$2/3Al_2O_3\rightarrow 4/3Al+O_2; \quad \Delta_rG=+966\$ kJ/mol.The potential difference needed for electrolytic reduction of \$Al_2O_3\$ at \$500^o\$C is at least:

The ionic reactions are,\$\dfrac{2}{3}\times(2Al^{3+})+4e^{-}\rightarrow \dfrac43Al\$\$\dfrac{2}{3}\times(3O^{2-})\rightarrow O_2+4e^{-}\$Thus the number of electrons transferred, \$n=4\$\$\Delta F=-nFe=-4\times96500\times E\$ \$966\times10^{3}=-4\times96500\times E\$\$\implies E=-\dfrac{966\times10^{3}}{4\times96500}=-2.5V\$

The standard reduction potentials \$E^{\circleddash}\$ for the half reactions are follows :\$Zn\, \rightarrow\, Zn^{2+}\, +\, 2e^{-};\, \quad\, E^{\circleddash}\, =\, +0.76\, V\$\$Fe\, \rightarrow\, Fe^{2+}\, +\, 2e^{-};\, \quad\, E^{\circleddash}\, =\, 0.41\, V\$The EMF for the cell reaction \$Fe^{2+}\, Zn\, \rightarrow\, Zn^{2+}\, +\, Fe\$ is:

\$Fe^{2+}\, +\, Zn\, \rightarrow\, Zn^{2+}\, +\, Fe\$\$Zn\, \rightarrow\, Zn^{2+}\, +\, 2e^-;\, \quad\, E^{\circleddash}\, =\, +0.76\, V\$\$Fe\, \rightarrow\, Fe^{2+}\, +\, 2e^-;\, \quad\, E^{\circleddash}\, =\, +0.41\, V\$These are oxidation potentials.Reduction potentials are equal and opposite. Fe forms cathode and Zn forms anode.\$E^{\circleddash}_{cell}\, =\, (E^{\circleddash}_{red})_c\, +\, (E^{\circleddash}_{oxid})_a\$= (-0.41 + 0.76) V= 0.35 V

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>>Class 12
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>> $Nernst Equation$

Nernst Equation

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Nernst Equation and Calculation of Emf

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Q: Calculate the emf of the cell. \$Zn | Zn^{2+} (0.001 M) || Cu^{2+} (0.1 M) | Cu\$The standard potential of \$Cu/Cu^{2+}\$ half-cell is +0.34 and \$Zn/Zn^{2+}\$ is -0.76 V.

Given \$Zn|Zn^{2+}(0.001M)||Cu^{2+}(0.1M)|Cu\$Overall cell reaction:\$Zn\longrightarrow { Zn }^{ 2+ }+2{ e }^{ - }\\ \underline { { Cu }^{ 2+ }+2{ e }^{ - }\longrightarrow Cu\quad \quad \quad } \\ \underline { Zn+{ Cu }^{ 2+ }\longrightarrow { Zn }^{ 2+ }+Cu } \$\$E_{cell}^{o}=\$standard reduction potential of cathode + standard oxidation potential of anode\$E_{cell}^{o}=\$0.34 to 0.76 V\$E_{cell}^{o}=\$1.1 V\${ K }_{ C }=\cfrac { \left[ { Zn }^{ 2+ } \right] }{ \left[ { Cu }^{ 2+ } \right] } =\cfrac { 10^{ -3 } }{ 10^{ -1 } } =10^{ -2 }\$EMF of the cell at any electrode concentration is:\$E=E^{ o }-\cfrac { 0.059 }{ n } \log \left( { K }_{ C } \right) =1.1-\cfrac { 0.059 }{ 2 } \log \left( 10^{ -2 } \right) =1.1-\cfrac { 0.059 }{ 2 } \times (2)=1.1-0.059=1.041V\$

Q: Calculate the emf of the following cell at \$298 K\$?\$Fe(s)|Fe^{2+}(0.001\ M)||H^{+}(1M)|H_{2}(g) (1\ bar), Pt(s)\$(Given: \$E^{\circ}_{cell} =+ 0.44\ V\$)

\$Fe(s)+2H^+ +(aq)\leftrightarrow Fe^{+2}(aq) + H_2(g)\$Nernst equation\$E_{cell}=E_{cell}^0 - \dfrac{0.0591}{n}log\,k_c\$There are two electron are transfering so that n=2\$E^0_{cell}=E^0_{right}-E^0_{left}\$\$E^0_{cell}=0-(-0.44)V\$\$E^0_{cell}=+0.44V\$Put the value we get Concentration of Solid substance is 1 always so that \$[Fe]=[H_2]=1\$\$\Rightarrow E_{cell}=0.44-\dfrac{0.0591}{2}log\dfrac{[Fe^{2+}]}{[H^+]^2}\$\$\Rightarrow = 0.44-\dfrac{0.0591}{2}log\left(\dfrac{0.001}{1^2}\right)\$\$\Rightarrow - E_{cell} = 0.44-\dfrac{0.0591}{2}log (10^{-3})\$\$\Rightarrow E_{cell}=0.44-\dfrac{0.0591}{2}(-3)\$\$\Rightarrow E_{cell}=0.44+0.089V\$\$\Rightarrow E_{cell}=0.53V\$

Q: Given the standard electrode potentials, \$K^{+}/K = -2.93\ V, Ag^{+}/ Ag = 0.80\ V,\$ \$Hg^{2+}/ Hg = 0.79\ V,\ Mg^{2+}/ Mg = -2.37V, Cr^{3+}/ Cr = -0.74\ V\$Arrange these metals in their increasing order of reducing power.

The increasing order of reducing power is as follows:\$\displaystyle Ag^+| Ag < Hg^{2+}| Hg < Cr^{3+}| Cr < Mg^{2+}| Mg < K^{+}| K\$Metal with most positive standard electrode potential has the least reducing power.Metal with most negative standard electrode potential has maximum reducing power.

Q: Calculate the emf of the cell\$Cr|Cr^{3+}(0.1\ M)||Fe^{2+} (0.01\ M)| Fe\$(Given: \$E_{Cr^{3+}/ Cr}^{\circ} = -0.75\ volt; E_{Fe^{2+}/ Fe}^{\circ} = -0.45\ volt)\$.

\$Cr|Cr^{3+}(0.1M)||Fe^{2+}(0.01M)|Fe\$\$E^{\circ}_{Cr^{2+}|Cr}=0.75V;\quad E^{\circ}_{Fe^{2+}/Fe}=-0.45 V\$\$2Cr^{3+}+3Fe\longrightarrow 3Fe^{2+}+2Cr\$\$\therefore\$ \$E_{cell}=E^{\circ}_{cell}-\cfrac{RT}{nF}ln Q\\=1.2-\cfrac{8.314\times 298}{6\times 96500}ln \cfrac{{(0.01)}^3}{{(0.1)}^2}\\=1.2-0.00428\times ln(10^{-4})\\=1.2-0.009855\times \log (10^{-4})\\=1.2+0.0394\\=1.2394\$

Nernst Equation

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Relation between EMF, Gibbs Free Energy, and Reaction Constant

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Numerical on Nernst Equation

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Nernst Equation and Calculation of Emf

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Electrochemical Cell and Gibbs Energy of the Reaction

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Calculate the emf of the cell
$C r ∣ C r^{3 +} (0.1 M) ∣ ∣ F e^{2 +} (0.01 M) ∣ F e$
(Given: $E_{C r^{3 +} / C r}^{\circ} = - 0.75 v o l t; E_{F e^{2 +} / F e}^{\circ} = - 0.45 v o l t)$ .

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Nernst Equation and Calculation of Emf

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Nernst Equation and Calculation of Emf

Electrochemical Cell and Gibbs Energy of the Reaction

Equilibrium Constant from Nernst Equation

Calculate the emf of the cell. Zn∣Zn2+(0.001M)∣∣Cu2+(0.1M)∣Cu The standard potential of Cu/Cu2+ half-cell is +0.34 and Zn/Zn2+ is -0.76 V.

The equilibrium constant of the reaction; Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s) Eo=0.46V at 298K

Represent the cell in which the following reaction takes place Mg(s)+2Ag+(0.0001M)→Mg2+(0.130M)+2Ag(s) Calculate its E(cell)​ifE(cell)o​=3.17V

Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd (ii) Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s) Calculate the Δr​G⊖ and equilibrium constant of the reactions

Calculate the emf of the following cell at 298K? Fe(s)∣Fe2+(0.001 M)∣∣H+(1M)∣H2​(g)(1 bar),Pt(s) (Given: Ecell∘​=+0.44 V)

Given the standard electrode potentials, K+/K=−2.93 V,Ag+/Ag=0.80 V, Hg2+/Hg=0.79 V, Mg2+/Mg=−2.37V,Cr3+/Cr=−0.74 V Arrange these metals in their increasing order of reducing power.

The Gibbs energy for the decomposition of Al2​O3​ at 500 oC is as follows: 2/3Al2​O3​→4/3Al+O2​;Δr​G=+966 kJ/mol. The potential difference needed for electrolytic reduction of Al2​O3​ at 500oC is at least:

The standard reduction potentials E⊝ for the half reactions are follows : Zn→Zn2++2e−;E⊝=+0.76V Fe→Fe2++2e−;E⊝=0.41V The EMF for the cell reaction Fe2+Zn→Zn2++Fe is:

Calculate the emf of the cell Cr∣Cr3+(0.1 M)∣∣Fe2+(0.01 M)∣Fe (Given: ECr3+/Cr∘​=−0.75 volt;EFe2+/Fe∘​=−0.45 volt).

Calculate the emf of the cell. $Z n ∣ Z n^{2 +} (0.001 M) ∣ ∣ C u^{2 +} (0.1 M) ∣ C u$
The standard potential of $C u / C u^{2 +}$ half-cell is +0.34 and $Z n / Z n^{2 +}$ is -0.76 V.

The equilibrium constant of the reaction;
$C u (s) + 2 A g^{+} (a q) ⟶ C u^{2 +} (a q) + 2 A g (s)$
$E^{o} = 0.46 V$ at $298 K$

Represent the cell in which the following reaction takes place $M g (s) + 2 A g^{+} (0.0001 M) \to M g^{2 +} (0.130 M) + 2 A g (s)$
Calculate its $E_{(c e l l)} if E_{(c e l l)}^{o} = 3.17 V$

Calculate the emf of the following cell at $298 K$ ?
$F e (s) ∣ F e^{2 +} (0.001 M) ∣ ∣ H^{+} (1 M) ∣ H_{2} (g) (1 b a r), P t (s)$
(Given: $E_{c e l l}^{\circ} = + 0.44 V$ )

Given the standard electrode potentials, $K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V,$ $H g^{2 +} / H g = 0.79 V, M g^{2 +} / M g = - 2.37 V, C r^{3 +} / C r = - 0.74 V$

Arrange these metals in their increasing order of reducing power.

The Gibbs energy for the decomposition of $A l_{2} O_{3}$ at $500$ $^{o} C$ is as follows:
$2 / 3 A l_{2} O_{3} \to 4 / 3 A l + O_{2}; Δ_{r} G = + 966$ kJ/mol.
The potential difference needed for electrolytic reduction of $A l_{2} O_{3}$ at $50 0^{o}$ C is at least:

The standard reduction potentials $E^{⊝}$ for the half reactions are follows :

$Z n \to Z n^{2 +} + 2 e^{-}; E^{⊝} = + 0.76 V$
$F e \to F e^{2 +} + 2 e^{-}; E^{⊝} = 0.41 V$

The EMF for the cell reaction $F e^{2 +} Z n \to Z n^{2 +} + F e$ is:

Calculate the emf of the cell
$C r ∣ C r^{3 +} (0.1 M) ∣ ∣ F e^{2 +} (0.01 M) ∣ F e$
(Given: $E_{C r^{3 +} / C r}^{\circ} = - 0.75 v o l t; E_{F e^{2 +} / F e}^{\circ} = - 0.45 v o l t)$ .