Determine the standard reduction potential for the half reaction:
${C l}_{2} + {2 e}^{-} ⟶ 2 {C l}^{-}$
Given: ${P t}^{2} + 2 C l ⟶ P t + {C l}_{2}, E_{c e l l}^{0} = - 0.15 V$
${P t}^{2} + {2 e}^{-} ⟶ P t, E^{0} = 1.20 V$

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Solution

Anode : ${2 C l}^{(-)} \to {C l}_{2} + {2 e}^{-}$
Cathode : ${P t}^{2 +} + 2 e^{-} \to P t$
$∴$ $E_{c e l l} = E_{R} - E_{L}$
$- 0.15 = E_{{P t}^{2 +} / P t} - E_{{C l}_{2} / {C l}^{(-)}}$
$- 0.15 = 1.20 - E$
$E = 1.20 + 0.15$
$E = 1.35 V$

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