Nernst Equation

Get insights from 14 questions on Nernst Equation, answered by students, alumni, and experts. You may also ask and answer any question you like about Nernst Equation

Follow Ask Question
14

Questions

0

Discussions

0

Active Users

0

Followers

New answer posted

12 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Ecell = 0 – log2

= – 0.03 (0.3)

 = – 0.009

 = – 9 * 10–3 V

x = 9

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

H2 (g) + Cu2+ (aq) →  2H+ (aq) + Cu (s)

  ∴ E c e l l = E c e l l 0 − 2 . 3 0 3 R T n f l o g Q

0.31 = 0.34 - 0 . 0 6 2 l o g [ H + ] C u 2 +  

  [ C u 2 + ] = 1 0 − 7 ( ? [ H + ] = 1 0 − 3 )

x = 7

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

H2 (g) + Cu2+ (aq) →  2H+ (aq) + Cu (s)

∴ E c e l l = E c e l l 0 − 2 . 3 0 3 R T n f l o g Q  

0.31 = 0.34 -  0 . 0 6 2 l o g [ H + ] C u 2 +  

[ C u 2 + ] = 1 0 − 7 ( ? [ H + ] = 1 0 − 3 )  

x = 7

New answer posted

a year ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

Ecell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 * 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

New answer posted

a year ago

0 Follower 17 Views

R
Raj Pandey

Contributor-Level 9

Benzyl amine (0.1 mole) reacts with 3 equivalents of CH? Br to form Benzyl trimethyl ammonium bromide.

Given 23g of Benzyl trimethyl ammonium bromide (molar mass 230 g/mol ), which is 0.1 mol.

Therefore, moles of CH? Br = 0.3 = 3 x 10? ¹. The value of n is 3.

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

The reaction shown is:
Anisole diazonium chloride (A) + Ethanol → Anisole + Acetaldehyde (X) + HCl + N? (Y)

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The reactions in the Solvay process are:

NaCl + H? O + NH? + CO? → NH? Cl + NaHCO?

2NaHCO? → Na? CO? + CO? + H? O

2NH? Cl + Ca (OH)? → CaCl? + 2NH? + H? O

CaCl? is a byproduct of the process.

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

1.52

E = E 0 - 0.0591 4 l o g ? H + 4

E = 1.23 + 0.0591 * p H

E = 1.23 + 0.0591 * ( 5 )

E = 1.52

New answer posted

a year ago

0 Follower 12 Views

R
Raj Pandey

Contributor-Level 9

Fe? ³ + e? → Fe? ² E° = 0.77V
Zn (s) → Zn? ² + 2e? ; E° = 0.76V
Cell reaction: 2Fe? ³ + Zn → 2Fe? ² + Zn? ² E°cell = 1.53V
Ecell = E°cell - (0.059/2)log ( [Zn? ²] [Fe? ²]²/ [Fe? ³]²)
1.5 = 1.53 - (0.06/2)log (1 * [Fe? ²]²/ [Fe? ³]²)
-0.03 = -0.03 log ( [Fe? ²]/ [Fe? ³])²
1 = log ( [Fe? ²]/ [Fe? ³])² => [Fe? ²]/ [Fe? ³] = 10
Let total iron = T. [Fe? ³] + [Fe? ²] = T. [Fe? ³] + 10 [Fe? ³] = T. 11 [Fe? ³] = T.
fraction of Fe? ³ = [Fe? ³]/T = 1/11 ≈ 0.09
This solution seems to differ from the image. Let's follow the image's steps.
log ( [Fe? ²]/ [Fe? ³])² = 1 => ( [Fe? ²]/ [Fe? ³])² = 10
[Fe? ²]/ [Fe?

...more

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 67k Colleges
  • 1.2k Exams
  • 716k Reviews
  • 1850k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.