In every electrolytic and galvanic (voltaic) cell the anode is the electrode at

Question

In every electrolytic and galvanic (voltaic) cell the anode is the electrode at which oxidation occurs. Which attracts cations. At which electrons are supplied to the solution. At which reduction occurs. Aqueous solutions of fluorides are difficult to oxidize at the anode of an electrolytic cell because the aqueous solutions of fluorides are nonconducting. It is impossible to find the proper material from which to build the electrodes. The fluorides are not very soluble. In salts the fluoride ion has a plus charge. Oxygen is released from water in preferences to fluorine. How many coulombs of electricity are required to completely convert 0.340 g of AgNO_3 into metallic Ag? 19.3 96.5 193 386 In the electrolysis of aqueous cupric sulfate using inert electrodes, passage through the cell of 96, 500 C will liberate 31.7 g of copper. 16.0 g of oxygen. 63.5 g of copper. 32.0 g of oxygen. What would be the E^0 value in volts for a zinc-silver galvanic cell? 0.76-0.80 0.76-(2 times 0.80) 0.76 + 0.80 0.76 + (2 times 0.80) What is true in the galvanic cell in which the overall reaction is The Br_2|Br^- electrode is the anode. Cations ions migrate towards the cathode. An increase in concentration of Mg^2+ would increase the cell voltage. The electrons enter the magnesium electrode from the external circuit. Calculate the potential of a voltaic cell in which the |Zn^2+| is 0.10 M and [Cu^2+] is 0.010 M (T = 25 degree C). 0.38 V 0.44 V 1.07 V 1.13V A solution of CdSO_4 is electrolyzed between inert electrodes How many hours must a current of 1.75 A flow to deposit 11.8 g of cadmium? 0.51 1.26 3.22 5.18

Explanation / Answer

1. Anode electrode is,

(A) where oxidation occurs

2. Aqueous fluoride is difficult to oxidize as,

(E) O2 is released from water

3. Coulombs of electrocity needed

(D)

4. grams of copper released

(B) 63.5 g copper

5. Eo value for the cell

(C) 0.80 + 0.76

6. In the given cell

(B) cations migrate towards cathode

7. Using Nernst equation

E = (0.34 + 0.76) - 0.0592/2 log(0.1/0.01)

(C) 1.07

8. hours of current needed to pass to deposit Cd

= 11.8 x 2 x 96500/112.4 x 60 x 60

(C) 3.22

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