Question

"The e.m.f. of the following cell at 298 K is 0.1745 V
Fe (s) / Fe²⁺ (0.1 M) // H+ (x M)/ H₂ (g) (1 bar)/ Pt (s)
Given : E⁰ (Fe²⁺/ Fe) = -0.44V
Calculate the H⁺ ions concentration of the solution at the electrode where hydrogen is being produced.

Answer 1

Answer:

Explanation:

Refer the below attachment.

Answer 2

The concentration of hydrogen ion is $6.24\times 10^{19}$.

Step by step explanation:

The given electrode reaction is as follows.

$\bold{Fe(s) + 2H^{+}\rightarrow Fe^{2+}(aq)+2e^{-}+H_{2}(g)}$

From the above reaction half cell reaction of Iron is as follows.

$\bold{Fe(s)\rightarrow Fe^{2+}(aq)+2e^{-}}$

By applying the Nernst equation to the above reaction we can calculate the hydrogen ion concentration.

Nernst equation is as follows.

$\bold{E_{cell}=E_{cell}^{o}-\frac{0.0591}{n}log(\frac{Fe^{2+}}{H^{+}})}$...............................(1)

From the given,

$\bold{E_{cell}=0.1745V}$

$\bold{E_{cell}^{o}=-0.44V}$

Molarity of $\bold{Fe^{2+}}$  = 0.1 M

Substitute the all given values in the equation (1)

$0.1745V=-0.44V-\frac{0.591}{n}log(\frac{0.1}{[H^{+}]})$

$-20.795=log\frac{0.1}{[H^{+}]}$

$log[H^{+}]=-19.795$

$[H^{+}]=6.24\times 10^{19}$

Therefore, hydrogen ion concentration is $\bold{6.24\times10^{19}}$.