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Electrochemistry

Question

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Write the Nernst equation and calculate the emf of the following cell at 298 K:
Pt(s) | Br₂ (l) Br^–(0.01M) || H⁺ (0.03 M) | H₂(g) (1 bar) | Pt(s)
Given E°_Br2/Br– = + 1.08 V
Br₂/ Br^–

Solution

The net reaction is 2Br^–(aq) + 2H⁺(aq) → H₂(g) + Br₂(l)
The Nernst equation is

E_{cell} = E °_{cell} - \frac{0.059}{2} \log [\frac{1}{{[H^{+}]}^{2} {[{Br}^{-}]}^{2}}] E °_{cell} = 0 - (+ 1.08) = 1.08

∴

E_{cell} = - 1.08 - \frac{0.059}{2} \log [\frac{1}{[0.03]^{2} [0.01]^{2}}]

E_{cell} = - 1.08 - 0.21 E_{cell} = - 1.29 V .

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