Q: Calculate EMF of Zn\mid _{(0.01M)}^{Zn^{2+}}||_{(0.1M)}^{Ag^{+}}|Ag E_{Zn/Zn^{+2}}^{0}=0.762VE_{Ag^{+}/Ag}^{0}=+0.80V

2Ag^+\rightarrow Zn^{2+}+2Ag E_{cell}=E^{\circ}_{cell}-\frac{RT}{nf}log \frac{[Zn^{2+}]}{[Ag^+]^2} E_{cell}=0.8+0.726-\frac{0.059}{2}log \left [ \frac{0.01}{(0.1)^2} \right ] E_{cell}=1.56 \therefore E_{cell}=1.56

Q: For hydrogen-oxygen fuel cell, the cell reaction is: 2{ H }_{ 2 }(g)+{ O }_{ 2 }(g)\longrightarrow 2{ H }_{ 2 }O(l) If \Delta { G }_{ f }^{ o }({ H }_{ 2 }O)=-237.2\ kJ{ mol }^{ -1 } , then emf of this cell is:

Gibbs free energy, \Delta { G }^{ o }=-nF{ E }^{ o } \therefore { E }^{ o }=\cfrac { \Delta { G }^{ o } }{ nF } =\cfrac { 237.2\times 1000J }{ 2\times 96500 } =1.23V[\because n=2]

Q: At 25^{\circ}C , the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl_{2} are 248\times 10^{-4}, 126\times 10^{-4} and 280\times 10^{-4} Sm^{2} mol^{-1} respectively, \lambda_{m}^{\circ}Ba(OH)_{2} in Sm^{2} mol^{-1} is

BaCl_{2} + 2NaOH\rightarrow Ba(OH)_{2} + 2NaCl \lambda^{\infty}_{m\ Ba(OH)_{2}} = \lambda^{\infty}_{m\ BaCl_{2}} + 2\lambda^{\infty}_{m\ NaOH} - 2\lambda^{\infty}_{m\ NaCl} = 280 \times 10^{-4} + 2\times 248\times 10^{-4} - 2 \times 126\times 10^{-4} = (280 + 496 - 252) \times 10^{-4} = 524\times 10^{-4} Sm^{2} mol^{-1} .

Question 1

Calculate EMF of  Zn\mid _{(0.01M)}^{Zn^{2+}}||_{(0.1M)}^{Ag^{+}}|Ag E_{Zn/Zn^{+2}}^{0}=0.762VE_{Ag^{+}/Ag}^{0}=+0.80V

Accepted Answer

2Ag^+ightarrow Zn^{2+}+2Ag E_{cell}=E^{\circ}_{cell}-\frac{RT}{nf}log \frac{[Zn^{2+}]}{[Ag^+]^2} E_{cell}=0.8+0.726-\frac{0.059}{2}log \left [ \frac{0.01}{(0.1)^2} ight ] E_{cell}=1.56 	herefore E_{cell}=1.56

Question 2

For a reaction,  A(s) + 2B^{+} \rightarrow A^{2+} + 2B ,  K_c  has been found to be  10^{12} . The  E^{o}_{cell}  is:

Accepted Answer

As we know, E\, =\, E^{\ominus}\, -\, \displaystyle \frac{0.059}{2}log\, K At equilibrium , E = 0   	herefore E^{o}\, =\, \displaystyle \frac{0.059}{2}log\, K\,=\, \displaystyle \frac{0.059}{2}log\, 10^{12}   \, =\, \displaystyle \frac{0.059}{2}\, 	imes\, 12\, =\, 0.354 V

Question 3

For hydrogen-oxygen fuel cell, the cell reaction is:

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l)

If

Δ G_{f}^{o} (H_{2} O) = - 237.2 k J m o l^{- 1}

, then emf of this cell is:

Accepted Answer

Gibbs free energy,  \Delta { G }^{ o }=-nF{ E }^{ o } 	herefore { E }^{ o }=\cfrac { \Delta { G }^{ o } }{ nF }                =\cfrac { 237.2	imes 1000J }{ 2	imes 96500 }                =1.23V[\because n=2]

Question 4

If the values of  \wedge_{\infty}  of  NH_{4}Cl, NaOH  and  NaCl  are  130, 217  and  109\ ohm^{-1} \cdot cm^{2}\cdot equiv^{-1}  respectively, the  \wedge_{\infty}  of  NH_{4}OH  in  ohm^{-1} \cdot cm^{2} \cdot equiv^{-1}  is:

Accepted Answer

\displaystyle \wedge_{\infty (NH_4OH)} = \wedge_{\infty (NH_4Cl)} + \wedge_{\infty (NaOH)} - \wedge_{\infty (NaCl)}   \displaystyle  \wedge_{\infty (NH_4OH)} = 130 + 217 - 109    \displaystyle  \wedge_{\infty (NH_4OH) }= 238 \: ohm^{-1} \cdot cm^{2} \cdot equiv^{-1}

Question 5

At  25^{\circ}C , the molar conductances at infinite dilution for the strong electrolytes  NaOH, NaCl  and  BaCl_{2}  are  248	imes 10^{-4}, 126	imes 10^{-4}  and  280	imes 10^{-4} Sm^{2} mol^{-1}  respectively,  \lambda_{m}^{\circ}Ba(OH)_{2}  in  Sm^{2} mol^{-1}  is

Accepted Answer

BaCl_{2} + 2NaOHightarrow Ba(OH)_{2} + 2NaCl \lambda^{\infty}_{m\&#10; Ba(OH)_{2}} = \lambda^{\infty}_{m\ BaCl_{2}} + &#10;2\lambda^{\infty}_{m\ NaOH} - 2\lambda^{\infty}_{m\ NaCl} = 280 	imes 10^{-4} + 2	imes 248	imes 10^{-4} - 2 	imes 126	imes 10^{-4} = (280 + 496 - 252) 	imes 10^{-4} = 524	imes 10^{-4} Sm^{2} mol^{-1} .

Question 6

The number of electrons participating in the electrode reaction when one atomic weight of a bivalent metal was deposited at the cathode:

Accepted Answer

Let, Mg ^{2+}  be the bivalent metal ion.Now,At cathode:  Mg^{2+}+2e^- ightarrow Mg(s) So, 2 moles of electrons is required to obtain 1 mol of metal atoms.So, number of electrons required  =2	imes 6.02	imes 10^{23}=12.04	imes 10^{23} Option C is correct.

Question 7

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:

Accepted Answer

MgBr_2  has   Mg^{2+}  ions. Mg^{2+ }    + 2e => Mg2 moles of electron deposits 1 mole of Mg.1 mole of electron = 1 Faraday3 moles of the electron will yield = 3/2 moles of Mg.1 mole of Mg= 24 gm.therefore 3/2 moles of Mg= 36 gm. Option C is correct.

Electrochemistry