Question

For the cell process:

\[\ce{Sn(s) + Pb^{+2}(aq) -> Pb(s) + Sn^{+2}(aq)}\]

The ratio of Pb⁺² to Sn⁺² ion concentration for spontaneity is ______.

Given: `"E"_("Sn"^(2+)//"Sn")^0` = − 0.136 V, `"E"_("Pb"^(2+)//"Pb")^0` = − 0.126 V

Options

• 0.197

• 0.225

• 0.374

• 0.458

Answer 1

For the cell process:

\[\ce{Sn(s) + Pb^{+2}(aq) -> Pb(s) + Sn^{+2}(aq)}\]

The ratio of Pb⁺² to Sn⁺² ion concentration for spontaneity is 0.458.

Explanation:

\[\ce{Sn(s) + Pb^{+2}(aq) -> Pb(s) + Sn^{+2}(aq)}\]

E⁰ = – 0.126 + 0 0.136 = 0.01 V

= – 2.303 × 8.31 × 298 log K

= – 2 × 96500 × 0.01

K = 2.18

∴ `"Pb"^(2+)/"Sn"^(2+) = 1/2.18` = 0.458