Question

What will be the emf for the given cell
Pt | H₂ (P₁) | H⁺ (aq) | | H₂ (P₂) | Pt
(a) RT/F logₑ P₁/P₂
(b) RT/2F logₑ P₁/P₂
(c) RT/F logₑ P₂/P₁ (d) None of these.

Answer 1

Explanation:

for the given cell

Pt | H₂ (P₁) | H⁺ (aq) | | H₂ (P₂) | Pt is (a) RT/F logₑ P₁/P₂

Answer 2

The emf for the given cell  Pt | H₂ (P₁) | H⁺ (aq) | | H₂ (P₂) | Pt RT/2F logₑ P₁/P₂.

Explanation:

Electromotive force

• The two electrodes of voltaic cell and galvanic cell is the potential difference to the maxima.
• The quantity relates elements to release or acquire electrons.
• This is the standard reduction potentials for the $E^ °$ as two half cells.
• Against the SHE, it is measured using the reduction potentials.

The anodic reaction is  H2(P1) → $2H^+$.

The cathodic reaction is $2H^+$→ H2(P2)

Ecathode = −RT/2F{ln(P2)/[H+]2}

Eanode = −RT/2F{ln([H+]2)/P1}

Emf = Eanode + Ecathode

Emf = −RT/2F{ln(P2)/[H+]2} - −RT/2F{ln([H+]2)/P1}

Emf  =−RT/2F{ln(P2)/P1}

Emf =RT/2F{ln(P1)/P2}

• Hence the cell is Pt | H₂ (P₁) | H⁺ (aq) | | H₂ (P₂) | Pt RT/2F logₑ P₁/P₂.

To learn more;

1. https://brainly.in/question/8828792
2. https://brainly.in/question/9849214