The correct representation of Nernst’s equation is

Question

The correct representation of Nernst’s equation is (A) EMn+/M=EoMn+M+(n/0.0591)log(Mn+) (B) EMn+/M=EoMn+M-(n/0.0591)log(Mn+) (C) EMn+/M=EoMn+M+(0.591/n)log(Mn+) (D) None of the above

Tardigrade Admin · Accepted Answer

E=Eo-(RT/nF)In (1/[Mn+]);E=Eo+(RT/nF)ln [Mn+] E=Eo+(2.303 RT/nF)log [Mn+] Substituting the value of R, T (298K) and F we get E=Eo+(0.0591/n)log(Mn+)

Q. The correct representation of Nernst’s equation is

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} + \frac{n}{0.0591} l o g (M^{n +})$

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} - \frac{n}{0.0591} l o g (M^{n +})$

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} + \frac{0.591}{n} l o g (M^{n +})$

None of the above

$E = E^{o} - \frac{RT}{n F} I n \frac{1}{[ M ^{n +} ]}; E = E^{o} + \frac{RT}{n F} l n [M^{n +}]$
$E = E^{o} + \frac{2.303 RT}{n F} l o g [M^{n +}]$
Substituting the value of $R, T (298 K)$ and $F$ we get
$E = E^{o} + \frac{0.0591}{n} l o g (M^{n +})$

Q. The correct representation of Nernst’s equation is

EMn+/M​=EMn+Mo​+0.0591n​log(Mn+)

EMn+/M​=EMn+Mo​−0.0591n​log(Mn+)

EMn+/M​=EMn+Mo​+n0.591​log(Mn+)

None of the above

E=Eo−nFRT​In[Mn+]1​;E=Eo+nFRT​ln[Mn+] E=Eo+nF2.303RT​log[Mn+] Substituting the value of R,T(298K) and F we get E=Eo+n0.0591​log(Mn+)

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} + \frac{n}{0.0591} l o g (M^{n +})$

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} - \frac{n}{0.0591} l o g (M^{n +})$

$E_{M^{n +} / M} = E_{M^{n +} M}^{o} + \frac{0.591}{n} l o g (M^{n +})$

$E = E^{o} - \frac{RT}{n F} I n \frac{1}{[ M ^{n +} ]}; E = E^{o} + \frac{RT}{n F} l n [M^{n +}]$
$E = E^{o} + \frac{2.303 RT}{n F} l o g [M^{n +}]$
Substituting the value of $R, T (298 K)$ and $F$ we get
$E = E^{o} + \frac{0.0591}{n} l o g (M^{n +})$