At 298 K, a 1 litre solution containing 10 mmol of Cr2O72– and 100 mmol of Cr3+ shows a pH of 3.0. Given: Cr2O72– → Cr3+ , Eº = 1.33V and The potential for the half cell reaction is x × 10

Engineering

Chemistry

Nernst Equation

Question

At 298 K, a 1 litre solution containing 10 mmol of Cr₂O₇^2– and 100 mmol of Cr³⁺ shows a pH of 3.0. Given: Cr₂O₇^2– → Cr³⁺ , Eº = 1.33V and $\frac{(2.303 RT)}{F} = 0.059 V$

The potential for the half cell reaction is x × 10^–3 V. The value of x is

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CrO₇^3– + 14H⁺ + 6e^– → 2Cr³⁺ +7H₂O

$\frac{(2.303 RT)}{F} = 0.059 V$

$E = 1.33 - \frac{0.059}{6} \times 42 = 0.917$

E = 917 × 10^–3

X = 917