MEDIUM

Earn 100

For the cell reaction, $Sn (s) + {Pb}^{+ 2} (aq) \to {Sn}^{2 +} (aq) + Pb (s),$
$E °_{{Sn}^{2 +} | Sn} = - 0.136 V, E °_{{Pb}^{2 +} | Pb} = - 0.126 V .$
Calculate the ratio of concentration of ${Pb}^{2 +}$ to ${Sn}^{2 +}$ ion at which the cell reaction will be reversed?

Important Questions on Electrochemistry

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:

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

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

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 :

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

For the electrochemical cell,

Mg(s) |{Mg}^{2+} (aq,1 M)| |{Cu}^{2+} (aq,1 M)| Cu (s)

the standard emf of the cell is

2 .70 V

300 K .

When the concentration of

{Mg}^{2+}

is changed to

x M,

the cell potential changes to

2 .67 V

300 K .

The value of

x

is ________.

(given,

\frac{F}{R} {=11500 K V}^{- 1},

where

F

is the Faraday constant and

R

is the gas constants,

ln (10) = 2 .30

)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

The photoelectric current from Na (work function, $w_{0} = 2.3 eV$ ) is stopped by the output voltage of the cell

$Pt (s) ‖ H_{2} (g, 1 bar) | HCl (aq \cdot, pH = 1) | AgCl (s) ∣ Ag (s)$
the $p H$ of aq. $HCl$ required to stop the photoelectric current from $K (w_{0} = 2.25 eV)$ , all other conditions remaining the same, is $\dots \dots \dots . \times 10^{- 2}$ (to the nearest integer).

Given $2.303 \frac{RT}{F} = 0.06 V; E_{AgCl / Ag / C l}^{0} = 0.22 V$

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

For an electrochemical cell

Sn (s) |{Sn}^{2 +} (aq, 1 M)| |{Pb}^{2 +} (aq, 1 M)| Pb (s)

the ratio

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

when this cell attains equilibrium is _______.
(Given:

E_{{Sn}^{2 +}| Sn}^{0} = - 0 .14 V, E_{{Pb}^{2 +}| Pb}^{0} = - 0 .13 V, \frac{2 .303 RT}{F} = 0 .06 \log 2.15 = \frac{1}{3}

)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

Find the electrode potential for the given half-cell reaction given below at $pH = 4$ .

$2 H_{2} O ⟶ O_{2} + 4 H^{+} + 4 e^{-}; E_{}^{°} = - 1.23 V$

(Given, $R = 8.314 J K^{- 1} {mol}^{- 1}$ , temperature $= 298 K$ , oxygen is under standard atmospheric pressure of $1 bar$ )

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 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

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

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

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)

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

$E_{1}, E_{2}$ 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>Nernst Equation

The

EMF

of a galvanic cell consisting of two hydrogen electrodes is

0.17 V

. If the solution of one of the electrodes has

[H^{+}] = 10^{- 3} M

, the

pH

at the other electrode is:

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

Given

E_{{Fe}^{+ 3 / {Fe}^{2}}} = + 0.76 V

and

E °_{I_{2} / I^{-}} = + 0.55 V

.
The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is

[\frac{2.303 RT}{F} = 0.06]

HARD

Chemistry>Physical Chemistry>Electrochemistry>Nernst Equation

What would be the electrode potential for the given half-cell reaction at

p H = 5 ?

2 H_{2} O \to O_{2} + 4 \overset{\oplus}{H} + 4 e^{-}; E_{r e d}^{0} = 1.23 V

(R = 8 .314 J {mol}^{- 1} K^{- 1}; Temp = 298 K; oxygen under standard atm . pressure of 1 bar)

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

)

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 -

For the cell reaction, Sn (s) + Pb+2 (aq) → Sn2+(aq) + Pb (s), E°Sn2+|Sn=-0.136 V, E°Pb2+|Pb=-0.126 V. Calculate the ratio of concentration of Pb2+ to Sn2+ ion at which the cell reaction will be reversed?

Important Questions on Electrochemistry

Learn and Prepare for any exam you want !

For the cell reaction, $Sn (s) + {Pb}^{+ 2} (aq) \to {Sn}^{2 +} (aq) + Pb (s),$
$E °_{{Sn}^{2 +} | Sn} = - 0.136 V, E °_{{Pb}^{2 +} | Pb} = - 0.126 V .$
Calculate the ratio of concentration of ${Pb}^{2 +}$ to ${Sn}^{2 +}$ ion at which the cell reaction will be reversed?