For the reaction taking place in the cell:
$Pt\left(s\right){H}_{2}\left(g\right){H}^{+}\left(aq\right)\left||{Ag}^{+}\left(aq\right)\left|Ag\left(s\right)\right\right)</p><p>$ {E}_{cell}^{°} = + 0 . 5332V . $</p><p>The value of$ {\Delta}_{f}{G}^{⊖} is ${kJmol}^{- 1}$ . (in nearest integer)

Question

For the reaction taking place in the cell: Pt(s) H _ 2 (g) H ^ + (aq)|| Ag ^ + (aq)|Ag(s)) E _ cell ^ ° = + 0 . 5332V . The value of Δ _ f G ^ ⊖ is kJmol ^ - 1 . (in nearest integer)

Quantrex Academy · Accepted Answer

Correct answer: 0

array ll Anode ⇒ (1)/(2) H _ 2 (g) ⟶ H ^ + (aq) + e ^ - Cathode ⇒ Ag ^ + + e ^ - ⟶ Ag ( s ) array (1)/(2) H _ 2 (g) + Ag ^ + (aq) n = 1 ⟶ H ^ + (aq) + Ag(s) Relation between Gibbs Free Energy and EMF of a Cell is given by Δ G ° = - nFE _ cell ^ ° = - 1 × 96500 × 0 . 5332 = 51453 . 8J mol ^ - 1 = 51 . 4538kJ mol ^ - 1

For the reaction taking place in the cell: Pt(s) H _ 2 (g) H ^ +...

Correct answer

Step-by-step solution

More from Electrochemistry