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 +

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For Daniell cell,
$Z n (s) + C u^{2 +} (a q) ⇌ C u (s) + Z n^{2 +} (a q)$
The reaction quotient Q is given by
$Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

OA = 1.10 volt
Where $E_{c e l l} = 1.1591$ volt, then

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Important Questions on Electrochemistry

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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 -

HARD

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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>Nernst Equation

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

)

EASY

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The standard reduction potentials (in V) of a few metal ion/metal electrodes are given below.

{Cr}^{3 +} / Cr = - 0 .74; {Cu}^{2 +} / Cu = + 0 .34; {Pb}^{2 +} / Pb = - 0 .13; {Ag}^{+} / Ag = + 0 .8 .

The reducing strength of the metals follows the order

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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:

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The standard cell potential for

Zn |{Zn}^{2 +}| |{Cu}^{2 +}| Cu

is 1.10 V. When the cell is completely discharged,

\log [{Zn}^{2 +}] / [{Cu}^{2 +}]

is closest to

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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

)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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:

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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>Nernst Equation

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

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

298 K

, the standard reduction potentials are

1.51 V

for

{M n O}_{4}^{-} | M n^{2 +}

1.36 V

for

C l_{2} | C l^{-}

1.07 V

for

B r_{2} | B r^{-}

0.54 V

for

I_{2} | I^{-}

. At

pH = 3

, permanganate is expected to oxidize:

(\frac{R T}{F} = 0.059)

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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:

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Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

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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>Nernst Equation

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

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

EASY

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

EASY

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

Among the following metals, the strongest reducing agent is

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

)

For Daniell cell, Zn(s)+Cu2+(aq)⇌Cu(s)+Zn2+(aq) The reaction quotient Q is given by Q=[Zn2+][Cu2+] OA = 1.10 volt Where Ecell=1.1591 volt, then

Important Questions on Electrochemistry

Learn and Prepare for any exam you want !

For Daniell cell,
$Z n (s) + C u^{2 +} (a q) ⇌ C u (s) + Z n^{2 +} (a q)$
The reaction quotient Q is given by
$Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

OA = 1.10 volt
Where $E_{c e l l} = 1.1591$ volt, then