Question

What will be the emf for the given cell:
$Pt$ ${H_2}$) (${P_1}$) ${H^ + }(aq)$ ${H_2}$(${P_2}$) $Pt$
 A. $\dfrac{{RT}}{F}\log \dfrac{{{P_1}}}{{{P_2}}}$
 B. $\dfrac{{RT}}{{2F}}\log \dfrac{{{P_1}}}{{{P_2}}}$
 C. $\dfrac{{RT}}{F}\log \dfrac{{{P_2}}}{{{P_1}}}$
 D. none of these

Answer 1

Hint: The Nernst equation is a formula that compares an atom's or ion's capacity to take up one or more electrons (reduction potential) under any conditions to that measured under standard conditions (standard reduction potentials) of $298K$ and one molar or atmospheric pressure.
Formula used:
Nernst equation:
$E = {E^0} - \dfrac{{2.303RT}}{{nF}}{\log _{10}}Q$
$E = $ Potential in volts
${E^0} = $Standard potential in volts
$R = $ Gas constant
$T = $ Temperature in Kelvin
$n = $ Number of moles of electrons transferred
$F = $ Faraday constant
$Q = $ Reaction quotient

Complete answer:
For the given cell, the reaction is as follows:
${H_2} \to 2{H^ + } + 2{e^ - }$
Here,$n = 2$
Using the Nernst equation given above,
For the cell's emf, the Nernst equation is:
$E_{cell}^0 = 0$
$n = 2$
Substituting the values, we get the following equation,
${E_{cell}} = 0 - \dfrac{{2.303RT}}{{2F}}\log \dfrac{{{P_2}}}{{{P_1}}}$
${E_{cell}} = \dfrac{{2.303RT}}{{2F}}\log \dfrac{{{P_1}}}{{{P_2}}}$

Hence, the correct option is B.

Note:
 In an extremely dilute solution, an ion's activity is close to infinity, hence it can be represented in terms of ion concentration. The ion concentration, on the other hand, is not equivalent to the ion activity in solutions with very high concentrations. In order to employ the Nernst equation in these situations, experimental measurements must be taken to determine the ion's true activity.