11. ⇒ (NEET 2013 )

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

(A) 0.118 V

(B) 1.18 V

(C) 0.059 V

(D) 0.59 V

Correct Answer is Option (D)

Oxidation half reaction is

H₂ $\to$ 2H⁺ + 2e^–

If pH = 10, [H⁺] = 10^–10

We know, Nernst Equation

E_cell = E°_cell - $\frac{0.0591}{2} \log \frac{{[H^{+}]}^{2}}{P_{H_{2}}}$

For hydrogen electrode E°_cell = 0

E_cell = - $\frac{0.0591}{2} \log \frac{{(10^{- 10})}^{2}}{1}$

= 0.0591log10¹⁰

E_cell = 0.591 V

12. ⇒ (AIPMT 2012 Mains )

The Gibb's energy for the decomposition of Al₂O₃ at 500^oC is as follows
$\frac{2}{3}$ Al₂O₃ $\to$ $\frac{4}{3}$ Al + O₂
$Δ$ _rG = +960 kJ mol^{$-$ 1}
The potential difference needed for the electrolytic reduction of aluminium oxide (Al₂O₃) at 500^oC is at least

(A) 4.5 V

(B) 3.0 V

(C) 2.5 V

(D) 5.0 V

Correct Answer is Option (C)

We know,

$Δ$ G^o = – nFE^o

$\frac{2}{3}$ Al₂O₃ $\to$ $\frac{4}{3}$ Al + O₂

Total number of Al atoms in Al₂O₃

= $\frac{2}{3} \times 2 = \frac{4}{3}$

Al³⁺ + 3e^– $\to$ Al

As 3e^– change occur for each Al atom

$∴$ n = $\frac{4}{3} \times 3 = 4$

E^o = - $\frac{Δ G^{\circ}}{n F}$

= - $\frac{960 \times 1000}{4 \times 96500}$

= - 2.5 V

13. ⇒ (AIPMT 2011 Prelims )

The electrode potentials for Cu²⁺_(aq) + e^$-$ $\to$ Cu⁺_(aq)
and Cu⁺_(aq) + e^$-$ $\to$ Cu_(s) are + 0.15 V and + 0.50 V respectively.

The value of E^o_cu²⁺/cu will be

(A) 0.500 V

(B) 0.325 V

(C) 0.650 V

(D) 0.150 V

Correct Answer is Option (B)

Cu²⁺_(aq) + e^$-$ $\to$ Cu⁺_(aq) ; E₁^o = 0.15 V

Cu⁺_(aq) + e^$-$ $\to$ Cu_(s) ; E₂^o = 0.50 V

Cu²⁺ + 2e^– $\to$ Cu ; E^o = ?

$Δ$ G^o = $Δ$ G₁° + $Δ$ G₂°

$\Rightarrow$ – nFE° = – n₁FE₁° – n₂FE₂°

$\Rightarrow$ E^o = $\frac{1 \times 0.15 + 1 \times 0.50}{2}$ = 0.325 V

14. ⇒ (AIPMT 2010 Prelims )

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25^oC. The value of standard Gibb's energy, $Δ$ G^o will be
(F = 96500 C mol^{$-$ 1})

(A) $-$ 89.0 kJ

(B) $-$ 89.0 J

(C) $-$ 44.5 kJ

(D) $-$ 98.0 kJ

Correct Answer is Option (A)

The cell reaction

Cu + 2Ag⁺ $\to$ Cu²⁺ + 2Ag

We know, $Δ$ G° = – nFE°_cell

= – 2 × 96500 × 0.46 = – 88780 J

= – 88.780 kJ = – 89 kJ

15. ⇒ (AIPMT 2009 )

Given :
(i)   Cu²⁺ + 2e^$-$ $\to$ Cu,  E^o = 0.337 V
(ii)  Cu²⁺ + e^$-$ $\to$ Cu⁺,  E^o = 0.153 V
Electrode potential, E^o for the reaction,
Cu⁺ + e^$-$ $\to$ Cu,  will be

(A) 0.90 V

(B) 0.30 V

(C) 0.38 V

(D) 0.52 V

Correct Answer is Option (D)

For the reaction,

Cu²⁺ + 2e^$-$ $\to$ Cu, E^o = 0.337 V

$Δ$ G^o = - nFE^o

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu²⁺ + e^$-$ $\to$ Cu⁺, E^o = 0.153 V

$Δ$ G^o = - nFE^o

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu²⁺ + e^$-$ $\to$ Cu⁺

$Δ$ G^o = –0.521 F = –nFE°

$\Rightarrow$ E° = 0.52 V

16. ⇒ (AIPMT 2008 )

Standard free energies of formation (in kJ/mol) at 298 K are $-$ 237.2, $-$ 394.4 and $-$ 8.2 for H₂O_(l), CO_2(g) and pentane _(g) respectively. The value of E^o_cell for the pentane-oxygen fuel cell is

(A) 1.0968 V

(B) 0.0968 V

(C) 1.968 V

(D) 2.0968 V

Correct Answer is Option (A)

At Anode:

C₅H₁₂ + 10H₂O $\to$ 5CO₂ + 32H⁺ + 32e^-

At Cathode:

8O₂ + 32H⁺ + 32e^- $\to$ 16H₂O
-------------------------------------------------
C₅H₁₂(g) + 8O₂(g) $\to$ 5CO₂(g) + 6H₂O(l)

$Δ$ G = 5× $Δ$ G_CO₂ + 6 $Δ$ G_(H₂O) – [ $Δ$ G_{(C₅H₁₂)} +8 × $Δ$ G_O₂ ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol^–1

= – 3387 × 10³ J mol^–1

$Δ$ G = - nFE^o_cell

From the overall equation we find n = 32

– 3387 × 10³ = -32 $\times$ 96500 $\times$ E^o_cell

E^o_cell = $\frac{- 3387 \times 10^{3}}{- 32 \times 96500}$ = 1.0968 V

17. ⇒ (AIPMT 2007 )

The equilibrium constant of the reaction:
Cu_(s) + 2Ag⁺_(aq) $\to$ Cu²⁺_(aq) + 2Ag_(s);
E^o = 0.46 V at 298 K is

(A) 2.0 $\times$ 10¹⁰

(B) 4.0 $\times$ 10¹⁰

(C) 4.0 $\times$ 10¹⁵

(D) 2.4 $\times$ 10¹⁰

Correct Answer is Option (C)

RT ln K = nFE°

ln K = $\frac{n F E^{\circ}}{R T}$

= $\frac{2 \times 0.46}{0.0591}$

$\Rightarrow$ K = 4 $\times$ 10¹⁵

18. ⇒ (AIPMT 2004 )

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25^oC. The equilibrium constant of the reaction would be

(A) 2.0 $\times$ 10¹¹

(B) 4.0 $\times$ 10¹²

(C) 1.0 $\times$ 10²

(D) 1.0 $\times$ 10¹⁰

Correct Answer is Option (D)

We know, from Nernst Equation

E_cell = E^o_cell - $\frac{2.303 R T}{n F} \log_{10} K$

At equilibrium E_cell = 0

$∴$ 0 = E^o_cell - $\frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log_{10} K$

$\Rightarrow$ 0 = 0.295 - $\frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log_{10} K$

$\Rightarrow$ 0.295 = $\frac{0.0591}{2} \log_{10} K$

$\Rightarrow$ $\log_{10} K$ = 10

$\Rightarrow$ K = 1 $\times$ 10¹⁰

19. ⇒ (AIPMT 2003 )

On the basis of the information available from the reaction,

4/3Al + O₂ $\to$ 2/3Al₂O₃, $Δ$ G = $-$ 827 kJ mol^{$-$ 1} of O₂,

the minimum e.m.f. required to carry out an electrolysis of Al₂O₃ is
(F = 96500 C mol^{$-$ 1})

(A) 2.14 V

(B) 4.28 V

(C) 6.42 V

(D) 8.56 V

Correct Answer is Option (A)

4/3Al + O₂ $\to$ 2/3Al₂O₃, $Δ$ G = $-$ 827 kJ mol^{$-$ 1}

For 1 mol of Al, n = 3

$∴$ For $\frac{4}{3}$ mol of Al, n = $3 \times \frac{4}{3} = 4$

As $Δ$ G = - nFE^o

$\Rightarrow$ – 827 × 10³ J = – 4 × E° × 96500

$\Rightarrow$ E^o = $\frac{827 \times 10^{3}}{4 \times 96500}$ = 2.14 V

20. ⇒ (AIPMT 2000 )

For the disproportionation of copper
2Cu⁺ $\to$ Cu²⁺ + Cu, E^o is
(Given E^o for Cu²⁺/Cu is 0.34 V and
E^o for Cu²⁺/Cu⁺ is 0.15 V.)

(A) 0.49 V

(B) $-$ 0.19 V

(C) 0.38 V

(D) $-$ 0.38 V

Correct Answer is Option (C)

Cu²⁺ + 2e^– $\to$ Cu; E°₁ = 0.34 V .....(1)

Cu²⁺ + e^– $\to$ Cu⁺ ; E°₂ = 0.15 V.....(2)

Cu⁺ + e^– $\to$ Cu; E°₃ = ? .....(3)

$∴$ $Δ$ G^o₁ = -2 $\times$ 0.34 $\times$ F

and $Δ$ G^o₂ = -1 $\times$ 0.15 $\times$ F

and $Δ$ G^o₃ = -1 $\times$ E°₃ $\times$ F

Also, $Δ$ G^o₁ = $Δ$ G^o₂ + $Δ$ G^o₃

$\Rightarrow$ -0.68F = -0.15F - E°₃ $\times$ F

$\Rightarrow$ E°₃ = 0 68 - 0 15 = 0 53 V

$∴$ E°_cell = 0.53 - 0.15 = 0.38 V

⇒Electrochemistry

Topic 2 : Nernst Equation 2