Question

P6C.2 Calculate the value of A_G®(H,0,1) at 298 K from the standard potential of the cell Pt(s)[H2(g)|HCl(aq)|O2(g)|Pt(s), El = +1.23 V.

Answer 1

Given cell : Pt (s) I H ₂ (g) I HCl (aq) IO ₂ (g) I Pt (s)

Consider reactions taking place at anode & cathode of given cell.

At anode: H ₂ (g) $\rightarrow$ 2 H ⁺ (aq) + 2 e ^-

At cathode : 1/2 O ₂ (g) + 2 H ⁺ (aq) + 2 e ^-  $\rightarrow$ H₂O (l)

Overall reaction : H ₂ (g) + 1/2 O ₂ (g) $\rightarrow$ H₂O (l)

In overall reaction, number of electrons transferred are 2.

The standard emf of above cell is 1.23 V

We have relation , $\Delta$ G ⁰ = - n F E ⁰ cell

Where, $\Delta$ G ⁰ is standard Gibbs energy of reaction, n is no. of electrons used in overall reaction , F is Faraday's constant & E ⁰ cell is standard emf of the cell.

$\therefore$$\Delta$G ⁰ = - 2 $\times$ 96485 C /mol $\times$ 1.23 V

$\Delta$G ⁰ = - 237.35 k J / mol

product of overall reaction is liquid water. Hence,Gibbs energy of reaction is the Gibbs energy of formation of liquid water.

ANSWER : $\Delta$ _fG ⁰ H₂O (l) = - 237.35 k J / mol