Question

The standard reduction potential for the half-cell reaction, $C{l}_2+2e^{-}\to 2C{l}^{-}$ will be:
$P{t}^{2+}+2C{l}^{-}\to Pt+C{l}_2,E_{cell}^o=-0.15$V;
$P{t}^{2+}+2e^{-}\to Pt,E^o=1.20$V

• A. $-1.35$V
• B. $+1.35$V
• C. $-1.05$V
• D. $+1.05$V

Answer 1

The correct option is B $+1.35$V

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, $P{t}^{2+}+2C{l}^{-}\to Pt+C{l}_2,E_{cell}^0=-0.15V$

On reversing the above reaction, $Pt+C{l}_2\to P{t}^{2+}+2C{l}^{-},E_{cell}^0=0.15V$

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

As we know the relation,

$E_{cell}^0=E_{cathode}^0-E_{anode}^0$

So we get,

$E_{C{l}_2/2C{l}^{-}}^0=0.15-(-1.20)=1.35V$

Hence, the correct option is $\text{B}$