Question
Calculate emf of the half-cell given below :
$${Pt(s)\left| H_{2}(\text{ }g,2\text{ }atm) \right|HCl(aq,0.02M) }{E_{H_{2}/H^{+}}^{\circ} = 0\text{ }V }$$(Given: $\frac{2.303RT}{F} = 0.059,log2 = 0.3010$ )
Options
Solution
Sol. [Key Concept: EMF of a half-cell via the Nernst equation]
$H_{2}(\text{ }g) \longrightarrow 2H^{\oplus}(aq) + 2e^{\ominus}$
$$\because E = E^{\circ} - \frac{2.303RT}{nF}log\frac{\left\lbrack H^{+} \right\rbrack^{2}}{P_{H_{2}}} $$$\left\lbrack H^{+} \right\rbrack = 0.02M$ (from HCl)
$${P_{H_{2}} = 2\text{ }atm }{\frac{2.303RT}{F} = 0.059 }{E = 0 - \frac{0.059}{2}log\frac{(0.02)^{2}}{2} = - 0.0295log\left( \frac{0.0004}{2} \right) }{E = - 0.0295log\left( 2 \times 10^{- 4} \right) = - 0.0295\left( log2 + log10^{- 4} \right) }{= - 0.0295\lbrack 0.301 - 4\rbrack = 0.109\text{ }V}$$
Create a free account to view solution
View Solution Free