1. Explain briefly, why upon the electrolysis of an aqueous solution of NaF the

Question

1. Explain briefly, why upon the electrolysis of an aqueous solution of NaF the product at the anode will be oxygen gas and at the cathode the product will be hydrogen gas.

2. How many grams of platinum will plate out at the cathode upon electrolysis of aqueous Pt(NO3)2 using a current of 4.60 Amp for 12.0 minutes. The atomic weight of Pt = 195.1.

3. A mixture of molten AlCl3 and MgBr2 is electrolyzed. Indicate what will be obtained at the anode _________________, and at the cathode ___________. Note: in using the attached table of reduction potential assume that the relative values of the reduction potentials for ions are the same in the molten state and in the aqueous state.

Can you please show work...I really need this and I don't understand!

Explanation / Answer

1)because the reduction potential of H+ is greater than Na+ so at cathode H+ undergo reduction to give H2 gas

and O2 oxidation potential is greater than F so the O2 can evolves at anode

2)

W = i*t*Z

    = 4.6* 720 *(195.1/2)

    = 323085.6 gm

3)

Al+3 + 3e-....> Al ...,   E = -1.662 V
Mg +2 +2e-.....> Mg ....   E = - 2.372 v

so Al Reduction potential is more so it can under go reduction at cathode

2Cl- + 2e- .....>Cl2.....E = - 1.358 v

2Br^- +2e- ......> Br2...E = -1.087 v

so Br- oxidation potential is more so it can under go oxidation togive Br2 gas at anode first

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