Question

A tin electrode in 0.015 M Sn (NO3)2 (aq) is connected to a hydrogen electrode in which the pressure of H2 is 1.0 bar. If the cell potential is 0.061 V at 250C, what is the pH of the electrolyte at the hydrogen electrode?

Answer 1

Answer:

The pH of the electrolyte at the hydrogen electrode is $2.24$·

Explanation:

Here, the tin electrode is connected with a standard hydrogen electrode which is known as a reference electrode·

The cell representation for the given cell is,

   $Sn\ /\ Sn^{2+}\ \ //\ \ H^+\ /\ H_2 \ \ (Pt)\\\\L\ H\ S \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R\ H\ S$

Here,

LHS is the anode and RHS is the cathode·

The oxidation will take place at the anode and the reduction will take place at the cathode·

The half cell reactions are,

At Anode,

 $Sn\ --->\ Sn^{2+}\ +\ 2e^-\ \ \ \ \ \ E^o_{(Sn^2^+/Sn)}\ =\ -0.14\ V$

At Cathode,

$2H^+\ +\ 2e^-\ --->\ H_2\ \ \ \ \ E^0_{(H^+/H)}\ =\ 0$

Since the reduction potential of $Sn^{2+}/Sn$ is lower than the reduction potential of hydrogen, we took the tin electrode as an anodic cell and the hydrogen electrode as a cathodic cell·

So the overall reaction of the cell becomes,

  $Sn\ +\ 2H^+\ --->\ Sn^{2+}\ +\ H_2$

Given that,

The concentration of  $Sn\ (NO_3)_2\ =\ 0.015\ M$

The partial pressure of $H_2\ =\ 1.0$ bar

The cell potential, $E_{cell}\ =\ 0.061\ V$

Temperature $=\ 25^{o} _{C} \ =\ 298\ K$

To find out the ph of the electrolyte at the hydrogen electrode first, we need to calculate the reaction quotient, Q from the Nernst equation·

The Nernst equation is,

$E_{cell}\ =\ E^0_{cell}\ -\ \frac{0.0592}{n} \ \ log\ Q$

where,

$E_{cell}$  $=$ cell potential

$E^0_{cell}$ $=$ standard reduction potential

n   $=$  No of electrons  $=\ 2\ e^-$

Q  $=$ reaction quotient

To calculate the $E^0_{cell}$,

We know that,

$E^0_{cell} \ =\ E^0_{(cathode)}\ -\ E^0_{(anode)}$

$E^0_{cell}\ =\ 0\ -\ (-0.14)$

Thus,

$E^0_{cell}\ =\ 0.14\ V$

On substituting the values we get,

⇒  $0.061\ =\ 0.140\ -\ \frac{0.0592}{2}\ \ log\ Q$

⇒  $-\ 0.079\ =\ -0.0296\ log \ Q$

⇒   $log \ Q\ =\ 2.67$

⇒    $Q\ =\ 467$

So, we get the value of Q as $467$ ·

To find out the value of ph we know that,

    pH   $=$  $- \ log\ {[H^+]}$

So now, we should find out the concentration of $H^+$ from the value of Q·

For that, the relation that connects the concentration of $H^+$ with the reaction quotient Q is,

    Q   $=$ $\frac{[Sn^{2+}]\ *\ P_{(H_2)}}{\ [H^+]^{2} }$

where,

$[Sn^{2+} ]\ =$ The concentration of $Sn^{2+}$ from the anodic half cell

$P_{(H_2)}\ =$  The partial pressure of hydrogen in the cathodic cell

$[H^{+} ]^{2} \ =$ The concentration of $H^+$ raised by its stoichiometric coefficient $2.$

On substituting the values,

⇒  $467$  $=$  $\frac{0.015\ *\ 1.0}{[H^+]^2}$

⇒  ${[H^+]^2}\ =\ \frac{0.015}{467}$

⇒  ${[H^+]}\ =\ 0.005667\ M$

As we know that,

pH   $=$  $-\ log\ {[H^+]}$

pH   $=\ \ -\ log\ [0.005667]$

pH   $=\ 2.24$

Therefore, the pH of the electrolyte at the hydrogen electrode is $2.24$·