Calculate emf of the following cell at 25 °C:

Fe | Fe²⁺(0.001 M) || H⁺ (0.01 M) | H₂ (g) (1 bar) | Pt(s)

E°(Fe²⁺ | Fe) = –0.44 V E°(H⁺ | H₂ ) = 0.00 V

Solution

For the given cell representation, the cell reaction will be Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g)

The standard emf of the cell will be
$straight E subscript cell superscript 0 space equals straight E to the power of 0 space straight H to the power of plus divided by straight H subscript 2 space rightwards arrow space straight E to the power of 0 space Fe to the power of 2 plus end exponent divided by Fe space rightwards double arrow space straight E subscript cell superscript space 0 end superscript space equals 0 minus left parenthesis negative 0.44 right parenthesis space equals 0.44 straight V The space Nernst space equation space for space the space cell space reaction space at space 25 to the power of 0 space straight C space will space be space straight E subscript cell space equals straight E subscript cell superscript 0 space minus fraction numerator 0.0591 over denominator 2 end fraction log space fraction numerator left square bracket Fe to the power of 2 plus end exponent right square bracket over denominator left square bracket straight H to the power of plus right square bracket to the power of 2 plus end exponent end fraction space equals space 0.44 minus fraction numerator 0.059 over denominator 2 end fraction space log space fraction numerator 0.001 over denominator left parenthesis 0.01 right parenthesis squared end fraction$

=0.44-0.02955(log10)

=0.44-0.02955(1)

=0.41045V 0.41V

The Nernst equation for the cell reaction at 25 º C will be

=0.44-0.02955(log10)

=0.44-0.02955(1)

=0.41045V 0.41V

Some More Questions From Electrochemistry Chapter

The molar conductivity of 0.025 mol L^–1methanoic acid is 46.1 S cm² mol^–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H⁺) = 349.6 S cm²mol^–1 and λ° (HCOO^– ) = 54.6 s cm² mol^–1.

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Electrochemistry

Calculate emf of the following cell at 25 °C:

Fe | Fe²⁺(0.001 M) || H⁺ (0.01 M) | H₂ (g) (1 bar) | Pt(s)

E°(Fe²⁺ | Fe) = –0.44 V E°(H⁺ | H₂ ) = 0.00 V

Some More Questions From Electrochemistry Chapter

How would you determine the standard electrode potential of the system Mg²⁺/Mg?

Can you store copper sulphate solutions in a Zinc pot?

Why does the conductivity of a solution decrease with dilution?

Suggest a way to determine the Λ°_m value of water.

The molar conductivity of 0.025 mol L^–1methanoic acid is 46.1 S cm² mol^–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H⁺) = 349.6 S cm²mol^–1 and λ° (HCOO^– ) = 54.6 s cm² mol^–1.

If the current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire?

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Electrochemistry

Calculate emf of the following cell at 25 °C: Fe | Fe2+(0.001 M) || H+ (0.01 M) | H2 (g) (1 bar) | Pt(s) E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2 ) = 0.00 V

Some More Questions From Electrochemistry Chapter

How would you determine the standard electrode potential of the system Mg2+/Mg?

Can you store copper sulphate solutions in a Zinc pot?

Why does the conductivity of a solution decrease with dilution?

Suggest a way to determine the Λ°m value of water.

The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H+) = 349.6 S cm2 mol–1 and λ° (HCOO– ) = 54.6 s cm2 mol–1.

If the current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Calculate emf of the following cell at 25 °C:

Fe | Fe²⁺(0.001 M) || H⁺ (0.01 M) | H₂ (g) (1 bar) | Pt(s)

E°(Fe²⁺ | Fe) = –0.44 V E°(H⁺ | H₂ ) = 0.00 V

How would you determine the standard electrode potential of the system Mg²⁺/Mg?

Suggest a way to determine the Λ°_m value of water.

The molar conductivity of 0.025 mol L^–1methanoic acid is 46.1 S cm² mol^–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H⁺) = 349.6 S cm²mol^–1 and λ° (HCOO^– ) = 54.6 s cm² mol^–1.