Question

Which one of the following has a potential more than zero?
(A) $Pt,(1/2){H_2}(1atm)|HCl(1M)$
(B) $Pt,(1/2){H_2}(1atm)|HCl(2M)$
(C) $Pt,(1/2){H_2}(1atm)|HCl(0.1M)$
(D) $Pt,(1/2){H_2}(1atm)|HCl(0.5M)$

Answer 1

Hint: The electrode represented here is the hydrogen electrode and for this electrode the standard electrode potential is zero (0) at 298 K.
Also use the Nernst equation to calculate the electrode potentials for these cells.

Formula used:
Here we have used the Nernst equation for a hydrogen electrode:
${E_{cell}} = {E^ \circ }_{cell} - (0.0591/n)\log (1/[{H^ + }])$
Since for a hydrogen electrode: ${E^ \circ }_{cell}$ = 0, the Nernst equation can simply be written as:
${E_{cell}} = 0 - (0.0591/n)\log (1/[{H^ + }])$

Complete answer:
First let us see what is this hydrogen electrode that has been represented in these options.
-Sometimes metals like gold or platinum are used as inert electrodes because they do not participate in a reaction but provide a surface for oxidation and reduction reactions and for the conduction of electrons. Due to this platinum is used in hydrogen electrodes.
-A standard hydrogen electrode is basically a redox electrode which acts as a base for the thermodynamic scale of oxidation and reduction potentials.

A hydrogen electrode is used as a reference electrode for comparison with other electrodes. Its standard electrode potential is zero at a temperature of 298K.
Which means that for hydrogen electrodes : ${E^ \circ }_{cell}$ = 0.
Reaction taking place at the half cell of hydrogen electrode is:
${H^ + }_{(aq)} + {e^ - } \to {[(1/2){H_2}]_{(g)}}$
It is represented as: $Pt(s)|{H_2}(g)|{H^ + }(aq)$
-Now let’s begin with calculating the electrode potential for all the 4 options:
-For option (A) $Pt,(1/2){H_2}(1atm)|HCl(1M)$: n=1
Using equation: ${E_{cell}} = 0 - (0.0591/n)\log (1/[{H^ + }])$
= 0 – (0.0591/1) log (1/1) (log 1= 0)
= 0 – 0.0591 (0)
= 0 V
For this cell: ${E_{cell}}$ = 0 V
-For option (B) $Pt,(1/2){H_2}(1atm)|HCl(2M)$ : n=1
${E_{cell}}$ = 0 – (0.0591/1) log (1/2)
=0 – 0.0591 log (0.5) ( log 0.5 = -0.301)
= 0 – 0.0591 (- 0.301)
= + 0.0178 V
For this cell: ${E_{cell}}$ = + 0.0178 V (which is greater than 0)
-For option (C) $Pt,(1/2){H_2}(1atm)|HCl(0.1M)$ : n=1
${E_{cell}}$= 0 – (0.0591/1) log (1/0.1)
= 0 – 0.0591 log 10 (log 10 = 1)
= (- 0.0591) V
For this cell: ${E_{cell}}$= (- 0.0591) V (which is lesser than 0)
-For option (D) $Pt,(1/2){H_2}(1atm)|HCl(0.5M)$ : n=1
${E_{cell}}$ = 0 – (0.0591/1) log (1/0.5)
= 0 – 0.0591 log 2 (log 2 = 0.301)
= (-0.0177891) V

For this cell: ${E_{cell}}$= (-0.0177891) V (which is also lesser than 0)
-We have calculated the electrode potential for all the cells now and we have seen that only 1 cell has a potential of more than zero, and that is the cell in option (B).
So, the correct option is: (B)$Pt,(1/2){H_2}(1atm)|HCl(2M)$.

Note: In such questions the most common mistake is that we don’t remember that the shown representation is that of a hydrogen electrode and for this electrode the standard electrode potential is zero (0). So, just remember this.