QJEEM22Chfb52: JEE Main 2022 Chemistry - Class 12 JEEM22Chfb52 Solution: For the reaction taking place in the cell : Pt (s)

# At anode, oxidation occur

H₂ $$\to$$ 2H⁺ + 2e^$$-$$ ....... (1)

# At cathode, reduction occur

2Ag⁺ + 2e^$$-$$ $$\to$$ 2Ag ...... (2)

Adding equation (1) and (2), we get n = 2, where n = cancelled out electron

Now,

$$\mathrm{\Delta }{G}^{\circ }=-nF{E}_{cell}^o$$

$$=-2\times 96500\times 0.5332$$

$$=-102907.6$$

$$=-102.9$$ kJ/mol

$$=-103$$ kJ/mol