For the disproportionation reaction 2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K, In K (where K is the equilibrium constant) is

Question

For the disproportionation reaction 2Cu⁺(aq) $\rightleftharpoons$ Cu(s) + Cu²⁺(aq) at 298 K, In K (where K is the equilibrium constant) is________ × 10^–1.

Given $\left(E_{Cu^{2+}/Cu^+}^o=0.16V,\ E_{Cu^+/Cu}^o=0.52V,\ {RT\over F}=0.025\right)$

Numerical Value Type

Comments

Please can u tell how 0.025/1,
And what is the value of n?

Answer 1

Ans. 144

Sol. $E_{cell}^0=E_{Cu^+/Cu}^0-E_{Cu^{2+}/Cu^{+1}}^0$

= 0.52 – 0.16

= 0.36 V

$E_{cell}^0={RT\over nF}\ln K_{eq}$

$0.36={0.025\over1}\ln K$

ln K = 14.4

= 144 × 10^–1

Ans. 144