Question

Find out E_cell of following electrochemical cell . (\(E_{Br_2/Br^-}^{^{\circ}} = 1.09V\))
1. – 1.32 V
2. – 1.03 V
3. – 1.09 V
4. – 1.15V

Answer 1

Correct Answer - Option 1 : – 1.32 V

2Br^–\(\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\) Br₂ + 2e^–

2H⁺ + 2e– \(\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\) H₂

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2Br^– + 2H⁺ \(\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\) H₂ + Br₂

E_Cell E_lsy =– 1.09 – \(\frac{{0.059}}{2}\) log \(\frac{1}{{{{({{10}^{ - 2}})}^2} \times {{(1 \times {{10}^{-2}})}^2}}}\)

= – 1.09 – \(\frac{{0.059}}{2}\) log 10⁸    

= – 1.09 – \(\frac{{0.059}}{2}\) × 8

 = – 1.32 V