{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Electrochemistry Homework Set Worked Out Solutions

# Oftheremaining

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: metalg metal The solution contains Pt2+, thus the plating reaction is: Pt2+(aq) + 2e‐ Pt(s) 0.500 A=0.500 C * 55.0 s= 27.5 C 1 s 27.5 C * 1 mol e‐ = 0.000285 mol e‐ 96,486 C 0.000285 mol e‐ * 1 mol Pt = 0.000143 mol Pt ‐ 2 mol e 0.000143 mol Pt * 195.08 g Pt = 0.0278 g = 27.8 mg 1 mol Pt 28. The solution contains Pt2+, thus the plating reaction is: 3+(aq) + 3e‐ Cr(s) Cr 5.00 A=5.00 C * (30.0 min* 60)= 9,000 C 1 s 9.000 C * 1 mol e‐ = 0.0933 mol e‐ 96,486 C 0.0933 mol e‐ * 1 mol Cr = 0.0311 mol Cr ‐ 3 mol e 0.0311 mol Cr * 52.00 g Cr = 1.62 g 1 mol Cr 29. For this problem, we need to work backwards to find the moles of vanadium oxidized and the moles of electrons transferred: 114 mg V=0.114 g V * 1 mol V = 0.00224 mol V 50.94 g V 650 C * 1 mol e‐ = 0.00674 mol e‐ 96,486 C Thus, 0.00674 moles of electrons were transferred to oxidize 0.00224 moles of vanadium, or: ‐=2.8 3 electrons transferred per vanadium atom oxidized 0.00674 mol e 0.00224 mol V oxidation state = +3 30. In order to liberate the most metal, we need the least amount of electrons required to reduce a mole of metal, or the smallest change in oxidation state. This corresponds to a lesser amount of electricity. All of the given choices involve a +10 change, except C, which involves a +20 change. Thus C can be eliminated. Next, we need the greatest molecular weight, so that 1.0 g of metal can be liberated with the least amount of electricity. Of the remaining choices, Ag has the greatest molecular weight, so answer choice D is correct. Dr. Fus Electrochemistry Problem Set Solutions Chemistry 123 31. CrO42‐(aq) + 8H+ + 6e‐ Cr(s) + 4H2O(l) 1.00 g Cr * 1 mol Cr = 0.0192 mol Cr 52.00 g Cr 0.0192 mol Cr * 6 mol e‐ = 0.115 mol e‐ 1 mol Cr 0.115 mol e‐ * 96,486 C = 11,133 C 1 mol e‐ 11,133 C * 1 s = 1,855.5 s = 30.9 min 6 C 32. Use the equation: Ecell=E°cell­ (0.0592/n)*logQ (*at 25 °C; n=mol e‐) For Sn2+ + 2Fe3+ Sn4+ + 2Fe2+, find Q: 4+][Fe2+]2= (0.10)(0.10)2= 0.008 Q=[Sn [Sn2+][Fe3+]2 (0.50)(0.50)2 Ecell= 0.617 – (0.0592/2)*log(0.008)= +0....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online