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

Answer 2

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M =  0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 =  0.76 - 0.0592/2logQ

0.660 - 0.76  =  - 0.0592/2logQ

-0.1 =  - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 =  [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92  × 10^-3/[H^+]^2

[H^+]^2 = 7.92  × 10^-3/2.39 × 10^3

[H^+] = 1.82  × 10^-3

pH = -log[H^+]

pH = -log[ 1.82  × 10^-3]

pH = 2.7

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