Question

The cell potential for the given cell at 298 K.

\[\ce{Pt | H2_{(g)} (1 bar) | H{^+_{(aq)}} || Cu{^{+2}_{(aq)}} | Cu_{(s)}}\] is 0.31 V.

The pH of the acidic solution is found to be 3 whereas the concentration of Cu⁺² is 10^−x M. The value of x is ______ (Nearest integer).

(Given, \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}} = 0.34 V}\] and \[\ce{\frac{2.303 RT}{F} = 0.06 V}\]).

Answer 1

The value of x is 7.

Explanation:

Given: The given cell is

\[\ce{Pt | H2_{(g)} (1 bar) | H{^+_{(aq)}} || Cu{^{+2}_{(aq)}} | Cu_{(s)}}\]

E_cell = 0.31 V

\[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = 0.34 V

pH = 3 ⇒ [H⁺] = 10⁻³ M

\[\ce{\frac{2.303 RT}{F} = 0.06 V}\]

We need to find x such that [Cu²⁺] = 10^−x

For the overall cell reaction:

\[\ce{H2_{(g)} + Cu{^{2+}{(aq)}} -> 2H{^+_{(aq)} + Cu_{(s)}}}\]

The Nernst equation is:

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.06}{2} log \frac{[H^{+}]^2}{[Cu^{2+}]}}\]

\[\ce{0.31 = 0.34 - 0.03 log \frac{10^{-6}}{10^{-x}}}\]

0.31 = 0.34 − 0.03 log(10^{x − 6})

0.31 = 0.34 − 0.03(x − 6)

0.31 = 0.34 − 0.03(x − 6)

0.03(x − 6) = 0.34 − 0.31

0.03(x − 6) = 0.03

⇒ x − 6 = 1

x = 7