The cell potential for Zn | Zn 2+( aq )|| Sn x +| Sn is 0.801 V at 298 K. The reaction quotient for the above reaction is 10-2. The number of electrons involved in the given electrochemical cell reaction is. .... (Given E Zn 2+ mid Zn 0=-0.763 V, E Sn x + mid sn 0=+0.008 V and .(2.303 RT / F )=0.06 V )

Question

The cell potential for Zn | Zn 2+( aq )|| Sn x +| Sn is 0.801 V at 298 K. The reaction quotient for the above reaction is 10-2. The number of electrons involved in the given electrochemical cell reaction is. .... (Given E Zn 2+ mid Zn 0=-0.763 V, E Sn x + mid sn 0=+0.008 V and .(2.303 RT / F )=0.06 V ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

E = E 0-(2.303 RT / nF ) log Q Here, E =+0.801 V , E 0=0.008-(-0.763) =+0.771 V ∴ 0.801=+0.771-(0.06/ n ) log 10-2 ⇒ n =4

E=E0−nF2.303RT​logQ Here, E=+0.801V,E0=0.008−(−0.763) =+0.771V ∴0.801=+0.771−n0.06​log10−2 ⇒n=4

$E = E^{0} - \frac{2.303 RT}{n F} lo g Q$
Here, $E = + 0.801 V, E^{0} = 0.008 - (- 0.763)$
$= + 0.771 V$
$∴ 0.801 = + 0.771 - \frac{0.06}{n} lo g 1 0^{- 2}$
$\Rightarrow n = 4$