Question

Explain nernst equation to determine the electrode potential of a cell

Answer 1

Hi mate.


Thanks for asking this question.


Here is your answer,




$nernst \: equation \: =$



The cell potential and electrode

potential depending upon temperature,

concentration of solute and partial pressure of

the gases.

The nernst equation is given as,

$e = {e}^{o} - \frac{rt}{nf} \: ln(q)$
Where,

$q = \frac{(product)}{(reactant)}$
$e = cell \: potential \\ {e}^{o} = standard \: cell \: potential \\ r = gas \: const. = 8.314j {k}^{ - 1} {mol}^{ - 1} \\ t = temperature = 298k$
$n = no. \: of \: moles \\ f = charge = 96500c$

We get a equation,

$e = {e}^{o} - \frac{0.0592}{n} \: \: log_{10}( \frac{product}{reactant} )$


#
(a) NERNST EQUATION FOR OXIDATION POTENTIAL OF ELECTRODE :


Consider a single electrode of metal dipped in its own salt solution. So, oxidation takes place.

$m \: > > > {m}^{ + n} \: + n {e}^{ - }$

$e = {e}^{o} - \frac{0.0592}{n} \: \: \: log_{10}( \frac{( {m}^{ + n} )}{m} )$

Where, (m) = 1

Then,

$e = {e}^{o} - \frac{0.0592}{n} \: \: log_{10}( {m}^{ + n} )$


#
(b) NERNST EQUATION FOR REDUCTION POTENTIAL AT ELECTRODE :


Consider a metal electrode dipped
in its own salt. Then, reduction takes place.


${m}^{ + n} + n {e}^{ - } > > > > m$

Then,

$e = {e}^{o} - \frac{0.0592}{n} \: \: log_{10}( \frac{(m)}{( {m}^{ +n} )} )$

Where, (m) = 1

Then,


$e = {e}^{o} - \frac{0.0592}{n} \: \: log_{10}( \frac{1}{ ({m}^{ + n} )} )$

$e = {e}^{o} + \frac{0.0592}{n} \: \: log_{10}( {m}^{ + n} )$



Hope it help you.



Thank you.



BE BRAINLY !!!!!!!!!!!!!

Answer 2

Answer:

The cell potential and electrode

potential depending upon temperature,

concentration of solute and partial pressure of

the gases.

The nernst equation is given as,

e = {e}^{o} - \frac{rt}{nf} \: ln(q)e=e

o

−

nf

rt

ln(q)

Where,

q = \frac{(product)}{(reactant)}q=

(reactant)

(product)

\begin{gathered}e = cell \: potential \\ {e}^{o} = standard \: cell \: potential \\ r = gas \: const. = 8.314j {k}^{ - 1} {mol}^{ - 1} \\ t = temperature = 298k \\ \end{gathered}

e=cellpotential

e

o

=standardcellpotential

r=gasconst.=8.314jk

−1

mol

−1

t=temperature=298k

\begin{gathered}n = no. \: of \: moles \\ f = charge = 96500c\end{gathered}

n=no.ofmoles

f=charge=96500c

We get a equation,

e = {e}^{o} - \frac{0.0592}{n} \: \: log_{10}( \frac{product}{reactant} )e=e

o

−

n

0.0592

log

10

(

reactant

product

)