The standard reduction potential the half-cell reaction- Cl-2-2e-rightarrow 2Cl- will be-Pt-2-2Cl-rightarrow Pt-Cl-2- E-o-cell-0-15V-Pt-2-2e-rightarrow Pt- E-o-1-20V-1-35V-1-35V-1-05V-1-05V

Question

Toppr Admin · Accepted Answer

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

The given reaction is, $P t^{2 +} + 2 C l^{-} \to P t + C l_{2}, E_{c e l l}^{0} = - 0.15 V$

On reversing the above reaction, $P t + C l_{2} \to P t^{2 +} + 2 C l^{-}, E_{c e l l}^{0} = 0.15 V$

Also given, $P t^{2 +} + 2 e^{-} \to P t, E_{c e l l}^{0} = 1.20 V$

As we know the relation,

$E_{c e l l}^{0} = E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$

So we get,

$E_{C l_{2} / 2 C l^{-}}^{0} = 0.15 - (- 1.20) = 1.35 V$

Hence, the correct option is $B$

The standard reduction potential for the half-cell reaction, Cl2+2e−→2Cl− will be:Pt2++2Cl−→Pt+Cl2,Eocell=−0.15V;Pt2++2e−→Pt,Eo=1.20V−1.35V+1.35V−1.05V+1.05V

The standard reduction potential for the half-cell reaction, $C l_{2} + 2 e^{-} \to 2 C l^{-}$ will be:
$P t^{2 +} + 2 C l^{-} \to P t + C l_{2}, E_{c e l l}^{o} = - 0.15$ V;
$P t^{2 +} + 2 e^{-} \to P t, E^{o} = 1.20$ V

$- 1.35$ V
$+ 1.35$ V
$- 1.05$ V
$+ 1.05$ V