Question

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

• A. $-1.35\text{V}$
• B. $+1.35\text{V}$
• C. $-1.05\text{V}$
• D. $+1.05\text{V}$

Answer 1

The correct option is B $+1.35\text{V}$

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,
$P{t}^{2+}+2C{l}^{-}\to Pt+C{l}_2,E_{\text{cell}}^o=-0.15V$
On reversing the above reaction.
$Pt+C{l}_2\to P{t}^{2+}+2C{l}^{-}{E}_{\text{cell}}^o=0.15V$
Also given, $P{t}^{2+}+2e^{-}\to Pt,E_{\text{cell}}^o=1.20V$
As we know the relation.
$E_{\text{cell}}^o=E_{\text{cathode}}^o-E_{\text{anode}}^o$
Since, $Pt$ is getting oxidised, so we get,
$E_{C{l}_2/2C{l}^{-}}^o=(0.15+1.20)V=1.35V$
Hence the correct option is (B)