If the standard electrode potential for a cell is $2V$ at $300K,$ the equilibrium constant \left(K\right) $for the reaction.<br /><br />$ Zn\left(s\right) $+ C{u}^{2 +}\left(aq\right)$ ⇌ Z{n}^{2 +}\left(aq\right) $+ Cu\left(s\right)$

at $300K$ is approximately:

\left(R = 8J{K}^{- 1}mo{l}^{- 1}, F = 96000Cmo{l}^{- 1}\right)

Question

If the standard electrode potential for a cell is

2V

at

300K,

the equilibrium constant \left(K\right)

for the reaction.<br /><br />

Zn\left(s\right)

+ C{u}^{2 +}\left(aq\right)

⇌ Z{n}^{2 +}\left(aq\right)

+ Cu\left(s\right)

at

300K

is approximately:

\left(R = 8J{K}^{- 1}mo{l}^{- 1}, F = 96000Cmo{l}^{- 1}\right)

Quantrex Academy · Accepted Answer

Correct answer: C. e ^ 160

∆ G = ∆ G ^ o + RTlnQ At equilibrium ∆ G = 0, Q = k 0 = ∆ G ^ o + RTlnK ∆ G ^ o = - RTlnK(∴ ∆ G ^ o = - nF E ^ o ) So - nF E ^ o = - RTlnk ′n′ factor for the above cell = 2 So - 2 × 96000 × 2 = - 8 × 300lnk lnk = 160 k = e ^ 160

If the standard electrode potential for a cell is 2V at 300K, the...

Options

Correct answer

Step-by-step solution

More from Electrochemistry