JEE PYQ: Electrochemistry Question 36
Question 36 - 2019 (11 Jan Shift 2)
Given the equilibrium constant $K_C = 10 \times 10^{15}$ for the reaction: $\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$ Calculate the $E^0_{\text{cell}}$ at 298 K. [$2.303\frac{RT}{F}$ at 298 K = 0.059]
(1) 0.04736 mV
(2) 0.4736 mV
(3) 0.4736 V
(4) 0.04736 V
Show Answer
Answer: (3) 0.4736 V
Solution
$E^0_{\text{cell}} = \frac{0.059}{2}\log 10^{16} = \frac{0.059}{2} \times 16 = 0.4736$ V.
BETA
