An acidified solution was electrolyzed using copper electrodes. a constant current of 1.18 a caused the anode to lose 0.584 g after 1.52 ✕ 103 s. (a) what is the gas produced at the cathode and what is its volume at stp? name of gas volume of gas webassign will check your answer for the correct number of significant figures. l (b) given that the charge of an electron is 1.6022 ✕ 10−19 c, calculate avogadro's number. assume that copper is oxidized to cu2+ ions.

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

2H⁺ +2e⁻ ⇌ H₂                    

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

d. Moles of H₂ formed

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

(b) Avogadro's number

(i) Moles of electrons transferred

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

(iv) Avogadro's number

first one is the correct answer

mark him as !