Question

A voltaic cell is set up with copper and hydrogen half-cells. Standard conditions are used in the copper half-cell, Cu2+ (aq, 1.00 M) | Cu (s). The hydrogen gas pressure is 1.00 bar. A value of 0.490 V is recorded for E Cell at 298 K. Determine the concentration of H+ and the pH of the solution.

Answer 1

Answer:

the concentration of H+ = 3 ×  $10^{-3}$   M and the pH = 2.6 of the solution

Explanation:

Based on reduction potentials, hydrogen is a better reducing agent than copper, therefore copper($Cu^{2+}$) is the cathode and hydrogen ($H_{2}$) is the anode.

Cathode reaction (reduction): $Cu^{2+}(aq) + 2e^{-}$  ⇒   Cu(s)

Anode reaction (oxidation) :  $H_{2}(g)$     ⇒    $2H^{+}(aq) + 2e^{-}$

net reaction:   $H_{2}(g) +$    $Cu^{2+}(aq)$   ⇒  $2H^{+}(aq) + Cu(s)$

$E_{0}cell = E_{0}cathode - E_{0}anode$

E cathode = 0.337 v

$E_{0}cell = + 0.337 - 0 = 0.337$

Q(reaction quotient) = $\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }$

for 2 electrons $Cu^{2+} = 1.00M$ but $H^{+}$ is unknown. we solve this using hernst equation.

$E = E^{0} -\frac{0.0257}{n}ln\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }$

$0.490 = 0.337 -\frac{0.0257}{2}ln\frac{[H^{+}]^{2} }{[1][1]}$

$ln{[H^{+}]^{2} } = -11.9$

$2ln{[H^{+}] } = -11.9$

$ln{[H^{+}] } = -5.95$

$[H^{+}] = 3* 10^{-3} M$

pH = 2.6