The relationship between standard reduction potential of a cell and equilibrium constant is shown by

Question

The relationship between standard reduction potential of a cell and equilibrium constant is shown by (A) E°cell=(n/0.059) log Kc (B) E°cell=(0.059/n) log Kc (C) E°cell= 0.059 nlogKc (D) E°cell=(log Kc/n)

Tardigrade Admin · Accepted Answer

At equilibrium E text cell =0. E c e l l=(0.059/Ω) log K c

Q. The relationship between standard reduction potential of a cell and equilibrium constant is shown by

$E_{ce ll}^{\circ} = \frac{n}{0.059} l o g K_{c}$

$E_{ce ll}^{\circ} = \frac{0.059}{n} l o g K_{c}$

$E_{ce ll}^{\circ} =$ 0.059 nlogK_c

$E_{ce ll}^{\circ} = \frac{l o g K _{c}}{n}$

At equilibrium $E_{cell} = 0$ .
$E_{ce ll} = \frac{0.059}{Ω} lo g K_{c}$

Q. The relationship between standard reduction potential of a cell and equilibrium constant is shown by

Ecell∘​=0.059n​logKc​

Ecell∘​=n0.059​logKc​

Ecell∘​= 0.059 nlogKc

Ecell∘​=nlogKc​​

At equilibrium Ecell ​=0. Ecell​=Ω0.059​logKc​

$E_{ce ll}^{\circ} = \frac{n}{0.059} l o g K_{c}$

$E_{ce ll}^{\circ} = \frac{0.059}{n} l o g K_{c}$

$E_{ce ll}^{\circ} =$ 0.059 nlogK_c

$E_{ce ll}^{\circ} = \frac{l o g K _{c}}{n}$

At equilibrium $E_{cell} = 0$ .
$E_{ce ll} = \frac{0.059}{Ω} lo g K_{c}$