Answer:
the concentration of H+ = 3 ×  [tex]10^{-3}[/tex]  M and the pH = 2.6 of the solution
Explanation:
Based on reduction potentials, hydrogen is a better reducing agent than copper, therefore copper([tex]Cu^{2+}[/tex]) is the cathode and hydrogen ([tex]H_{2}[/tex]) is the anode.
Cathode reaction (reduction): [tex]Cu^{2+}(aq) + 2e^{-}[/tex]  ⇒  Cu(s)
Anode reaction (oxidation) :  [tex]H_{2}(g)[/tex]   ⇒   [tex]2H^{+}(aq) + 2e^{-}[/tex]
net reaction:  [tex]H_{2}(g) +[/tex]   [tex]Cu^{2+}(aq)[/tex]  ⇒  [tex]2H^{+}(aq) + Cu(s)[/tex]
[tex]E_{0}cell = E_{0}cathode - E_{0}anode[/tex]
E cathode = 0.337 v
[tex]E_{0}cell = + 0.337 - 0 = 0.337[/tex]
Q(reaction quotient) = [tex]\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }[/tex]
for 2 electrons [tex]Cu^{2+} = 1.00M[/tex] but [tex]H^{+}[/tex] is unknown. we solve this using hernst equation.
[tex]E = E^{0} -\frac{0.0257}{n}ln\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }[/tex]
[tex]0.490 = 0.337 -\frac{0.0257}{2}ln\frac{[H^{+}]^{2} }{[1][1]}[/tex]
[tex]ln{[H^{+}]^{2} } = -11.9[/tex]
[tex]2ln{[H^{+}] } = -11.9[/tex]
[tex]ln{[H^{+}] } = -5.95[/tex]
[tex][H^{+}] = 3* 10^{-3} M[/tex]
pH = 2.6
