Which is the correct relationship between equilibrium constant and standard reduction potential of cell?

Question

Which is the correct relationship between equilibrium constant and standard reduction potential of cell? (A) Ec e l l0=0.059nlogKc (B) Ec e l l0=(0.059/n)logKc (C) Ec e l l0=(n/0.059)logKc (D) Ec e l l0=(l o g Kc/n)

Tardigrade Admin · Accepted Answer

The equilibrium constant and standard reduction potential of cell is given by Ec e l lo=(2.303 R T/n F)logK=(0.0591/n)logKcat298 K

Q. Which is the correct relationship between equilibrium constant and standard reduction potential of cell?

$E_{ce ll}^{0} = 0.059 n l o g K_{c}$

$E_{ce ll}^{0} = \frac{0.059}{n} l o g K_{c}$

$E_{ce ll}^{0} = \frac{n}{0.059} l o g K_{c}$

$E_{ce ll}^{0} = \frac{l o g K _{c}}{n}$

The equilibrium constant and standard reduction potential of cell is given by $E_{ce ll}^{o} = \frac{2.303 RT}{n F} l o g K = \frac{0.0591}{n} l o g K_{c} a t 298$ $K$

Q. Which is the correct relationship between equilibrium constant and standard reduction potential of cell?

Ecell0​=0.059nlogKc​

Ecell0​=n0.059​logKc​

Ecell0​=0.059n​logKc​

Ecell0​=nlogKc​​

The equilibrium constant and standard reduction potential of cell is given by Ecello​=nF2.303RT​logK=n0.0591​logKc​at298 K

$E_{ce ll}^{0} = 0.059 n l o g K_{c}$

$E_{ce ll}^{0} = \frac{0.059}{n} l o g K_{c}$

$E_{ce ll}^{0} = \frac{n}{0.059} l o g K_{c}$

$E_{ce ll}^{0} = \frac{l o g K _{c}}{n}$

The equilibrium constant and standard reduction potential of cell is given by $E_{ce ll}^{o} = \frac{2.303 RT}{n F} l o g K = \frac{0.0591}{n} l o g K_{c} a t 298$ $K$