Question

A Galvanic cell is consisting of standard concentrations: Cu|Cu2+||Ag+|Ag, is connected to a small bulb for 45.0 minutes. During that time the average current drawn by the bulb is 0.15 A. How many grams of copper will dissolve from the anode during this time?

Answer 1

Given :

The given Galvanic cell with standard concentrations is :

$Cu|Cu^{2+}||Ag^+|Ag$

Time for which this cell is connected to a bulb = 45 mins

Average current drawn by the bulb during this 45 mins = 0.15 A

To Find :

In this process how many grams of copper will dissolve from the anode ?

Solution :

Since we know that :

$Charge = current \times time$

Here the reaction involved is :

$Cu^{2+} + 2e^- \rightarrow Cu$

We can convert the time in seconds , so :

t = $45 \times 60$ sec

 = 2700 sec

So , the total charge = $0.15 \times 2700$ C

                                  = 405 C

Now we know that  96500 C = 1 F

So , 405 C = $\frac{1}{96500}\times 405$ F

                  = 0.0041 C

Now for converting 1 mole $Cu^{2+$ into Cu , 2F charge is required .

So , 0.0041 F can deposit $\frac{0.0041}{2}$ moles of Cu.

So , total moles of Cu deposit = $\frac{0.0041 }{2} = 0.00205$ moles

So , mass of Cu deposited = $0.00205 \times 63.5$ g

                                             = 0.1301 g

Hence , 0.1301 gm of copper will dissolve from the anode .