Question

Calculate the emf of half cell :
$Pt \begin{array}{l} Cl _{2}( g ) \\ (10 atm )\end{array} \mid \begin{array}{l} HCl \\ (0.1) M \end{array} E _{ Cl _{2} \mid Cl ^{-}}^{0}=1.36 V$

• A. 1.45 V
• B. 1.27 V
• C. -1.45 V
• D. -1.27 V

Answer 1

Answer: (A)

$Cl _{2} \rightarrow 2 Cl ^{-}+2 e ^{-}$

$E = E ^{\circ}-\frac{0.0591}{ n } \log _{10} \frac{\left[ Cl ^{-}\right]^{2}}{ P _{ Cl _{2}}}$

$1.36-0.059 / 2 \log \frac{(0.1)^{2}}{10}=1.36-0.0295 \times-3=1.45 V$