Question

JEE Mains question is in the image

Answer 1

For chemical reaction,

Cu(s) + Sn²⁺(aq ) ⇔Cu²⁺(aq) + Sn(s)

here [Cu²⁺] = [Sn²⁺] = 1M

given , E°_{Cu²⁺/Cu} = 0.34V and E°_{Sn²⁺/Sn}= -0.16V

To find : The Gibb's energy change ( in KJ)

first find standard electromotive force of cell, E°_(cell)

using E°_(cell) = E°_{Sn²⁺/Sn} - E°_{Cu²⁺/Cu}

= -0.16 - 0.34 = - 0.50 V

now Gibb's energy change, ∆G = -nFE°_(cell)

here n = no of electrons exchange = 2

F = 96500 C

= -2 × 96500 × -0.5

= 96500 J

= 96.5 kJ

Therefore the Gibb's energy change is 96.5 KJ