COMPASSLearning

Talk to us Register Login

CBSE Class 12 · 2015 · Delhi · Redox Reactions and ElectrochemistryShort Answer

Calculate the emf of the following cell at 25°C:

Fe ∣ F e^{2 +} (0.001 M) ∣∣ H^{+} (0.01 M) ∣ H_{2} (g) (1 bar) ∣ Pt (s)

E^{\circ} (F e^{2 +} ∣Fe) = - 0.44

V,

E^{\circ} (H^{+} ∣ H_{2}) = 0.00

V.

Answer: E_cell ≈ 0.41 V

Solution

Cell reaction: Fe + 2H

^{+} \to

Fe

^{2 +}

+ H

_{2}

,

n = 2

:

E_{ce l l}^{\circ} = 0.00 - (- 0.44) = 0.44 V

Nernst equation:

E_{ce l l} = 0.44 - \frac{0.059}{2} lo g \frac{[ F e ^{2 +} ] p _{H_{2}}}{[ H ^{+} ] ^{2}} = 0.44 - 0.0295 lo g \frac{1 0 ^{- 3}}{1 0 ^{- 4}}

= 0.44 - 0.0295

E_{ce l l} = 0.41 V

CBSE Class 12 Board · 2015 · Delhi · Subjective

Practise more: all CBSE board PYQs on Redox Reactions and Electrochemistry →

More board questions on Redox Reactions and Electrochemistry

Which of the following cells is used in the Apollo space programme?
CBSE 2024View solution →
Dilution affects both conductivity as well as molar conductivity. The effect of dilution on both is as follows:
CBSE 2024View solution →
State the following:
(a) Kohlrausch law of independent migration of ions
(b) Faraday's first law of electrolysis
CBSE 2024View solution →
The resistance of a conductivity cell filled with 0.2 mol L $^{- 1}$ KCl solution is 200 Ω. If the resistance of the same cell when filled with 0.05 mol L $^{- 1}$ KCl solution is 620 Ω, calculate the conductivity and molar conductivity of the 0.05 mol L $^{- 1}$ KCl solution. The cond …
CBSE 2024View solution →
Calculate the emf of the following cell:
$Ni (s) + 2A g^{+} (0.01 M) ⟶ N i^{2 +} (0.1 M) + 2Ag (s)$

Given: $E_{ce l l}^{\circ} = 1.05$ V, log 10 = 1.
CBSE 2024View solution →
Which one of the following cells was used in the Apollo space programme?
CBSE 2023View solution →