Question

The cell potential for the given cell at 298 K Pt | H₂ (g, 1 bar) | H⁺ (aq) | | Cu²⁺ (aq) | Cu(s) is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^-x M. The value of x is ______.

[Given: (\[\ce{E_{Cu^{2+}/Cu}}\]) = 0.34 V and `(2.303 " RT")/"F"` = 0.06 V]

Options

• 11

• 7

• 20

• 6

Answer 1

The cell potential for the given cell at 298 K Pt | H₂ (g, 1 bar) | H⁺ (aq) | | Cu²⁺ (aq) | Cu(s) is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^-x M. The value of x is 7.

Explanation:

Given cell:

Pt | H₂(g) | H⁺ (ag) | | Cu²⁺ (aq) | Cu (s)

Oxidation Half reaction:

\[\ce{H2 -> 2H+  +2e^-, E^0_{H°/H_2}}\] = 0 V

Reduction Half reaction:

\[\ce{Cu^{2+} + 2e^- -> Cu, E^0_{H°/H_2}}\] = 0.34 V

Net cell reaction:

\[\ce{H2 + Cu^{2+} -> 2H^+ + Cu}\]

\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]

\[\ce{E^0_{cell}}\] = (0.34 - 0) V = 0.34 V

Now, pH of the acidic solution = 3   ...(given)

pH = - log₁₀[H⁺]

3 = - log₁₀[H⁺]

⇒ [H⁺] = 10^-3 M

[Cu²⁺] = 10^-x M

E_cell = 0.31 V  (given)

Applying Nernst equation,

\[\ce{E_{cell} = E^0_{cell}}\] - `(2.303 " RT")/"nF" log  ["H"^+]^2/[["Cu"^(2+)]]`

`0.31 = 0.34 - 0.06/2 log  (10^-3)^2/10^(- x)`

`0.31 = 0.34 - 0.03 log (10^(- 6 - (- x)))`

- 0.03 = - 0.03 log `10^(- 6 + x)`

1 = log `10^(- 6 + x)`

1 = (- 6 + x) log 10

x = 1 + 6 = 7    ...[∵ log 10 = 1]

So, the value of x is 7.