Question

Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(P_H2)= 0.87atm.

Answer 1

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

$2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_{red}= 0.00 \ V}$

At the anode:

At oxidation reaction is:

$Zn \to Zn^{2+} +2e^- \ \ \ \mathbf{E^0_{ox} = 0.76 \ V}$

The overall equation for the reaction is:

$\mathbf{Zn + 2H^+ \to Zn^{2+} + H_2}$

The overall cell potential is:

$\mathbf{E^0_{cell}= E^0_{ox} + E^0_{red}}$

$\mathbf{E^0_{cell}= 0.76 \ V +0.00 \ V}$

$\mathbf{E^0_{cell}= 0.76\ V}$

Using the formula for the Nernst equation:

$E = E^0 - ( \dfrac{0.0591}{n})log (Q)$

where;

E = 0.66

(Zn^2+)=0.22 M

Then

$0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )$

$0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )$

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log  [H⁺]

4.118  = - 2  log  [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059