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)

43.452

step-by-step explanation:

284 times .153 = 43.452

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