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 +

EASY

Earn 100

The correct order of $E °_{M^{2 +} / M}$ values with the negative sign for the four successive elements $Cr, Mn, Fe$ and $Co$ is

50% studentsanswered this correctly

Important Questions on Electrochemistry

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

)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The standard cell voltage for the cell

Pb | {Pb}^{2 +} | | {Sn}^{2 +} | Sn

- 0.01 V .

If the cell is to exhibit

E_{cell} = 0

, the value of

\frac{[{Sn}^{2 +}]}{[{Pb}^{2 +}]}

should be antilog of

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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:

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

)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

EASY

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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 :

MEDIUM

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:

MEDIUM

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

MEDIUM

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

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The pressure of

H_{2}

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

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

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>Nernst Equation

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 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:

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The electrode potential of a hydrogen electrode at

pH = 10

MEDIUM

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}

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

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:

MEDIUM

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 -

The correct order of E°M2+/M values with the negative sign for the four successive elements Cr, Mn, Fe and Co is

Important Questions on Electrochemistry

Learn and Prepare for any exam you want !

The correct order of $E °_{M^{2 +} / M}$ values with the negative sign for the four successive elements $Cr, Mn, Fe$ and $Co$ is