Given: C o 3 + + e - → C o 2 + ; E ° = + 1.81 V P b 4 + + 2 e - → P b 2 + ; E ° = + 1.67 V C e 4 + + e - → C e 3 + ; E ° = + 1.61 V B i 3 + + 3 e - → B i ; E ° = + 0.20 V Oxidizing power of the species will increase in the order:

B i 3 + < C e 4 + < P b 4 + < C o 3 +

C o 3 + < C e 4 + < B i 3 + < P b 4 +

C o 3 + < P b 4 + < C e 3 + < B i 3 +

C e 4 + < P b 4 + < B i 3 + < C o 3 +

MEDIUM

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What will be value of log $K_{C}$ for the given electrochemical cell?

$C u |{C u}^{2 ⨁} (a q)| |A g^{⨁} (a q)| A g$

$[E_{c e l l}^{o} = 0.46 V]$

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Important Questions on Electrochemistry

EASY

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The standard electrode potentials

(E_{M^{+} / M}^{o})

of four metals

A, B, C

and

D

are

- 1 .2 V,

0 .6 V

0 .85 V

and

- 0 .76 V

, respectively. The sequence of deposition of metals on applying potential is

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

For the following electrochemical cell at

289 K,

Pt (s) | H_{2} (g, 1 bar) | H^{+} (aq, 1 M) | | M^{4 +} (aq .) {, M}^{2+} (aq .) | Pt (s)

E_{c e l l} = 0.092 V

when

\frac{[M^{2 +} (a q .)]}{[M^{4 +} (a q .)]} = 10^{x}

Given:

E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059

The value of

x

is -

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The metal that cannot be obtained by the electrolysis of an aqueous solution of its salt is

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

In the cell,

Pt (s) |H_{2} (g, 1 bar)| HCl (aq) |AgCl (s)| Ag (s) |Pt (s),

the cell potential is

0 .92 V

when a

10^{- 6}

molar

HCl

solution is used. The standard electrode potential of

Ag | AgCl | {Cl}^{-}

electrode is:

(Given,

\frac{2 .303 RT}{F} = 0 .06 V

298 K

)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The pressure of

H_{2}

required to make the potential of hydrogen electrode zero in pure water at 298 K is:

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Given:

C o^{3 +} + e^{-} \to C o^{2 +}; E ° = + 1.81 V

{P b}^{4 +} + {2 e}^{-} \to {P b}^{2 +}; E ° = + 1.67 V

C e^{4 +} + e^{-} \to C e^{3 +}; E ° = + 1.61 V

{B i}^{3 +} + {3 e}^{-} \to B i; E ° = + 0.20 V

Oxidizing power of the species will increase in the order:

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

In the electrochemical cell:

Zn |{ZnSO}_{4} (0 .01 M)| |{CuSO}_{4} (1 .0 M)| Cu,

the emf of this Daniel cell is

E_{1}

. When the concentration

{ZnSO}_{4}

is changed to

1 .0 M

and that of

{CuSO}_{4}

changed to

0 .01 M,

the emf changes to

E_{2}

. From the following, which one is the relationship between

E_{1}

and

E_{2}

? (Given,

\frac{RT}{F} = 0 .059

)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The electrode potential of a hydrogen electrode at

pH = 10

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The voltage of the cell consisting of

Li (s)

and

F_{2} (g)

electrodes is

5.92 V

at standard condition at

298 K

. What is the voltage if the electrolyte consists of

2 M LiF

.
Given that,

\ln 2 = 0 .693, R = 8 .314 {JK}^{- 1} {mol}^{- 1} F = 96500 C {mol}^{- 1}

EASY

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Which of the following ions does not liberate hydrogen gas on reaction with dilute acids?

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The correct representation of Nernst’s equation for half- cell reaction

{Cu}^{2 +} (aq) + e^{-} \to {Cu}^{+} (aq)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Given

E_{{Cr}^{3 +} / Cr}^{o} = - 0.74 {V;  E}_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{o} = 1.51 V

E_{{Cr}_{2} O_{7}^{2 -} / {Cr}^{3 +}}^{o} = 1.33 {V;   E}_{{Cl}_{2} | {Cl}^{-}}^{o} = 1.36 V

Based on the data given above, strongest oxidising agent will be:

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

250 mL

of a waste solution obtained from the workshop of a goldsmith contains

0.1 M {AgNO}_{3}

and

0.1 M AuCl

. The solution was electrolyzed at

2 V

by passing a current of

1 A

for

15

minutes. The metal/metals electrodeposited will be :

(E_{{Ag}^{+} / Ag}^{0} = 0.80 V, E_{{Au}^{+} / Au}^{0} = 1.69 V)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Given that,

$E_{O_{2} / H_{2} O}^{o} = + 1 .23 V;$
$E_{S_{2} O_{8}^{2 -} / {SO}_{4}^{2 -}}^{o} = 2 .05 V$
$E_{{Br}_{2} / {Br}^{-}}^{o} = + 1.09 V;$
$E_{{Au}^{3 +} / Au}^{o} = 1 .4 V$

The strongest oxidizing agent is

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

For the given cell; $Cu (s) |{Cu}^{2 +} (C_{1} M)| |{Cu}^{2 +} (C_{2} M)| Cu (s)$

change in Gibbs energy $(∆ G)$ is negative, it :

HARD

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

$E_{1}, E_{2}$ and $E_{3}$ are the emf of the following three galvanic cells respectively

(i)

Zn (s) |{Zn}^{2 +} (0.1 M)| |{Cu}^{2 +} (1 M)| Cu (s)

(ii) $Zn (s) |{Zn}^{2 +} (1 M)| |{Cu}^{2 +} (1 M)| Cu (s)$

(iii) $Zn (s) |{Zn}^{2 +} (1 M)| |{Cu}^{2 +} (0.1 M)| Cu (s)$

Which of the following is true?

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

The electrode potentials for ${Cu}^{2 +} (aqueous) + e^{-} ⟶ {Cu}^{+} (aqueous)$ and ${Cu}^{+} (aqueous) + e^{-} ⟶ Cu (s)$ are $+ 0.15 V$ and $+ 0.50 V,$ respectively. The value of $E °_{{Cu}^{2 +} / Cu}$ will be:

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

To find the standard potential of

M^{3 +} / M

electrode, the following cell is constituted:

Pt / M / M^{3 +} (0 .001 mol L^{- 1}) / {Ag}^{+} (0 .01 mol L^{- 1}) / Ag

The emf of the cell is found to be

0.421 volt

298 K

. The standard potential of half-reaction

M^{3 +} + 3 e^{-} \to M

298 K

will be:

(Given:

E_{\frac{{Ag}^{+}}{Ag}}^{⊝}

298 K = 0.80 volt

)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Relationship between Electrode Potential, Gibb's Free Energy and Equilibrium Constant

Consider the following reduction processes:

{Zn}^{2 +} + 2 e^{-} \to Zn (s); E^{o} = - 0 .76 V

{Ca}^{2 +} + 2 e^{-} \to Ca (s); E^{o} = - 2 .87 V

{Mg}^{2 +} + 2 e^{-} \to Mg (s); E^{o} = - 2 .36 V

{Ni}^{2 +} + 2 e^{-} \to Ni (s); E^{o} = - 0 .25 V

The reducing power of the metals increases in the order:

What will be value of log KC for the given electrochemical cell? CuCu2⨁aq Ag⨁(aq)Ag Ecello=0.46V

Important Questions on Electrochemistry

Learn and Prepare for any exam you want !

What will be value of log $K_{C}$ for the given electrochemical cell?

$C u |{C u}^{2 ⨁} (a q)| |A g^{⨁} (a q)| A g$

$[E_{c e l l}^{o} = 0.46 V]$