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 standard reduction potentials of Cu^{2+}\, |\, Cu and Cu^{2+}\, |\, Cu^{\bigoplus} are 0.337 V and 0.153 V, respectively. The standard electrode potential of Cu^{\bigoplus}\, |\, Cu half cell is:

Cu^{2+}\, +\, 2e^-\, \rightarrow\, Cu\, \quad\, ...(i) E^{\circleddash}_1 Cu^{2+}\, +\, e^-\, \rightarrow\, Cu^{\bigoplus}\, \quad\, ...(ii) E^{\circleddash}_2 Net equation (Cu^{\bigoplus}\, +\, e^-\, \rightarrow\, Cu)\, \quad\, ...(iii) E^{\circleddash}_3\, =\, \displaystyle \dfrac{n_1E^{\circleddash}_1\, -\, n_2E^{\circleddash}_2}{n_3}\, =\, \displaystyle \dfrac{2\, \times\, 0.337\, -\, 1\, \times\, 0.153}{1}\, =\, 0.521\, V Hence, option C is correct.

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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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: The standard electrode potential of Ag^{+}/Ag is +0.80\ V and of Cu^{2+}/ Cu is +0.34\ V . These electrodes are connected through a salt bridge and if:

Standard reduction potential of Ag^{+}/Ag=0.80V .Standard reduction potential of Cu^{2+}/Cu is less than Ag^{+}/Ag . \therefore\,Ag^{+}/Ag will get reduced at cathode. Cu^{2+}/Cu will get oxidized at anode. \therefore\,E_{cell}^{\circ}=E_{cathode}^{\circ}-E_{anode}^{\circ} E_{cell}^{\circ}=0.80-0.34=0.46V .

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 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

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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:

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Important Questions

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

The standard electrode potential of Ag+/Ag is +0.80 V and of Cu2+/Cu is +0.34 V. These electrodes are connected through a salt bridge and if:

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.

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)

The standard reduction potentials of Cu2+∣Cu and Cu2+∣Cu⨁ are 0.337 V and 0.153 V, respectively. The standard electrode potential of Cu⨁∣Cu half cell is:

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. $Z n ∣ Z n^{2 +} (0.001 M) ∣ ∣ C u^{2 +} (0.1 M) ∣ C u$
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 $E_{(c e l l)} if E_{(c e l l)}^{o} = 3.17 V$

The standard electrode potential of $A g^{+} / A g$ is $+ 0.80 V$ and of $C u^{2 +} / C u$ is $+ 0.34 V$ . These electrodes are connected through a salt bridge and if:

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.

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: $E_{c e l l}^{\circ} = + 0.44 V$ )

The standard reduction potentials of $C u^{2 +} ∣ C u$ and $C u^{2 +} ∣ C u^{⨁}$ are 0.337 V and 0.153 V, respectively. The standard electrode potential of $C u^{⨁} ∣ C u$ half cell is:

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: