Write the half reactions and the balanced equation for the galvanic cell Cu(s)| Cu2+(aq)|| Cu+(aq)| Cu(s). What is the smallest possible integer coefficient of Cu+(aq) in the combined balanced equation?

Answer and Explanation:

The galvanic cell is:

Cu(s)| Cu²⁺(aq)|| Cu⁺(aq)| Cu(s)

The first two species before the double bar (||) constitute the anodic half reaction (oxidation):

Cu(s) ⇒ Cu²⁺(aq) + 2 e-

The two species after the || constitute the cathodic half reaction (reduction):

Cu⁺(aq) + e- ⇒ Cu(s)

If we multiply the reduction half reaction by 2 (to obtain the same number of electrons than oxidation reaction) and then we add the two half reactions, we obtan the balanced equation:

Reduction (cathode) : 2Cu⁺(aq) +2 e- ⇒ 2Cu(s)

Oxidation (anode) :     Cu(s) ⇒ Cu²⁺(aq) + 2 e-

Total equation:         2Cu⁺(aq) + Cu(s) ⇒ 2Cu(s) + Cu²⁺(aq)

Cu(s) is in both reactants side and products side, so we cancel that in both opposite sides to obtain:

2Cu⁺(aq) ⇒ Cu(s) + Cu²⁺(aq)

If we divide the balanced equation into 2, the smallest possible integer coefficient for Cu⁺(aq) is 1:

Cu⁺(aq) ⇒ 1/2 Cu(s) + 1/2 Cu²⁺(aq)

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

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