I2 (s) I I-(0.1M) half cell is connected to a H+ (aq) |H2 (1bar)| Pt half cell and e.m.f. is found to be 0.7714 V. If EI2∣I-o=0.535V, find the pH of H+ |H2 half-cell.

Pt,1/2H2/H⊕/I(0.1)Θ/1/2I2,Pt1/2H2+1/2I2⟶H⊕+IΘE=E0−(0.0591/n)log⁡Q0.7714=(0.535−0)−(0.0591/1)log⁡H⊕I−H+=10−3,pH=3

I2 (s) I I-(0.1M) half cell is connected to a H+ (aq) |H2 (1bar)| Pt half cell and e.m.f. is found to be 0.7714 V. If EI2∣I-o=0.535V, find the pH of H+ |H2 half-cell.

Pt,1/2H2/H⊕/I(0.1)Θ/1/2I2,Pt1/2H2+1/2I2⟶H⊕+IΘE=E0−(0.0591/n)log⁡Q0.7714=(0.535−0)−(0.0591/1)log⁡H⊕I−H+=10−3,pH=3

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I₂ (s) I I^-(0.1M) half cell is connected to a H⁺ (aq) |H₂ (1bar)| Pt half cell and e.m.f. is found to be 0.7714 V. If $E_{I_{2} ∣ I^{-}}^{o} = 0.535 V$ , find the pH of H⁺ |H₂ half-cell.

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Detailed Solution

$\begin{array}{l} Pt, 1 / 2 H_{2} / H^{\oplus} / I_{(0 .1)}^{Θ} / 1 / 2 I_{2}, Pt \\ 1 / 2 H_{2} + 1 / 2 I_{2} ⟶ H^{\oplus} + I^{Θ} \\ E = E^{0} - (0 .0591 / n) \log Q \\ 0 .7714 = (0 .535 - 0) - (0 .0591 / 1) \log (H^{\oplus}) (I^{-}) \\ [H^{+}] = 10^{- 3}, pH = 3 \end{array}$