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Karl-Anthony is trying to plate gold onto his silver ring. He constructs an electrolyte cell using his ring as one of the electrodes. He runs this cell for 90.6 minutes at 213.8 mA. How many moles of electrons were transferred in this process?

Answer:

0.012 mole of electron

Explanation:

From the question given above, the following data were obtained:

Time (t) = 90.6 minutes

Current (I) = 213.8 mA

Number of mole of electrons =?

Next, we shall convert 90.6 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

90.6 mins = 90.6 × 60

90.6 mins = 5436 s

Next, we shall convert 213.8 mA to A. This can be obtained as follow:

1000 mA = 1 A

Therefore,

213.8 mA = 213.8 mA × 1 A / 1000 mA

213.8 mA = 0.2138 A

Next, we shall determine the quantity of electricity used in the process. This can be obtained as follow:

Time (t) = 5436 s

Current (I) = 0.2138 A

Quantity of electricity (Q) =?

Q = it

Q = 0.2138 × 5436

Q = 1162.2168 C

Next, the equation for the reaction.

Au⁺ + e —> Au

From the balanced equation above,

1 mole of electron was transferred.

Recall:

1 faraday = 96500 C = 1 e

Thus,

96500 C of electricity is needed to transfer 1 mole of electron.

Finally, we shall determine the number of mole electrons transferred by the application of 1162.2168 C of electricity. This can be obtained as follow:

96500 C of electricity is needed to transfer 1 mole of electron.

Therefore,

1162.2168 C of electricity will transfer = 1162.2168 / 96500 = 0.012 mole of electron

Thus, 0.012 mole of electron was transferred in the process.

Following are the calculation to the moles of electrons that transferred in the process:

Given

Current flows

time[tex]= 90.6\ min = 90.6 \times 60\ sec= 5436 \ sec\\\\[/tex]

For Step 1:

Calculating the Total charge flown in the given time:

For Step 2:

Charge on a single electron

So:

Following are the calculation to the charges of electrons:

Following are the calculation to the electrons transferred:

[tex]=\frac{(726.38\times 10^{19})}{ (6.023 \times 10^{23})}\\\\= 120.60\times 10^{-4}\ moles\\\\[/tex]

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