JEE PYQ: Electrochemistry Question 9
Question 9 - 2021 (26 Feb Shift 2)
EMF of the following cell at 298 K in V is $x \times 10^{-2}$. $\text{Zn}|\text{Zn}^{2+}(0.1\text{M})|\text{Ag}^+(0.01\text{M})|\text{Ag}$ The value of x is ____.
(Round off to the nearest integer) [Given: $E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V; $E^0_{\text{Ag}^+/\text{Ag}} = +0.80$ V; $\frac{2.303RT}{F} = 0.059$]
Show Answer
Answer: 147
Solution
$E^0 = 1.56$ V. $E = 1.56 - \frac{0.059}{2}\log\frac{0.1}{(0.01)^2} = 1.56 - 0.0885 = 1.4715$. $x = 147$.
BETA
