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

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Solution

Rules of balancing redox reaction :
(1) separate the redox reaction into half-reaction
(2) Balance the atom in each half reaction
a) Balance all atom except H and O
b) Balance the oxygen atom with $H_{2} O$
c) Balance the hydrogen atom with $H^{+}$
d) In a basic medium, add one $O H^{+}$ to each side
for every $H^{+}$
(3) Balance the change with $e^{-}$
(4) Make $e^{-}$ gain equivalent to election loss in the half-reaction
(5) Add the half reaction together
(6) Simplify the equation
(7)finally check the element and changes are balanced
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To find reduction potential for half reaction $(E^{\circ} c e l l)$
$c l_{2} + 2 e^{-} \to 2 c l^{-}$
given $e q^{n} p t^{2} + 2 c l^{-} \to p t + c l_{2}$ $E_{c e l l} = - 0.15 v . . . (1)$
$p t^{2} + 2 e^{-} \to p t$ $E = 1.20 V . . . (2)$
$e q^{n} (1) \Rightarrow 2 c l_{2} + p t \to p t^{2} + 2 c l^{-}$ $E_{c e l l} = 0.15 V . . . (3)$
Adding $e q^{n}$ (2) and $e q^{n} (3)$
$p t^{2} + 2 e^{-} \to p t$
$c l_{2} + p t \to p t^{2} + 2 c l^{-}$
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$c l_{2} + 2 e^{-} \to 2 c l^{-}$
$∴ E^{\circ} c e l l = E^{\circ} o x i d a t i o n + E^{\circ} r e d a c t i o n$
$= 0.15 + 1.20$
$∴ E^{\circ} c e l l = 1.35 v$

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