How many faradays is needed to \"plate out\" 100.0 g of chromium? (use table 41.

Question

How many faradays is needed to "plate out" 100.0 g of chromium? (use table 41.1 for charge) A. 5.8 B. 5.5 x 105 C. 1.8 x 10-6 The balanced overall equation for: Au3+(aq) + Br-(aq) Au(s) + Br2(g) A. 3Au3+(aq) + 6Br-(aq) 3Au(s) + 3 Br2(g) B. 3Au3+(aq) + 4Br-(aq) 3Au(s) + 2 Br2(g) C. 2Au3+(aq) + 6Br-(aq) 2Au(s) + 3 Br2(g) Calculate the Ecell of: Al(s) + Cu2+(aq)(0.010M) Al3+(aq)(1.0M) + Cu(s) (is this balanced?) A. 2.06 B. 1.94 C. -2.06 I realize I have more than 1 question however, I have posted some before with conflicting answers. I really do appreciate and will rate. Thank you again

Explanation / Answer

Pat A ) Faradays needed to plate out 100 g Cr = 100 g x 3/51.996 g/mol = 5.8 Faradays

Answer : A. 5.8

Part B) Overall balanced euation,

Answer,

C. 2Au3+(aq) + 6Br-(aq) ---> 2Au(s) + 3Br2(g)

Part C) Ecell = Eo - 0.0592/n logK

                     = (0.337 - (-1.66) - 0.0592/6 log(1/0.01^3)

                     = 1.94 V

Anser : B. 1.94

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