\"A voltaic cell is constructed from two half-cells; silver wire dipped in 50.0m

Question

"A voltaic cell is constructed from two half-cells; silver wire dipped in 50.0ml AgNO3 (aq) of unknown concentration and nickel wire dipped in 50.0ml 0.400M Ni(NO3)2 (aq). The measured voltage when first connected is 1.10V. The standard reduction potentials are 0.7994V for silver and -0.25V for nickel.

How much time would need to pass for the measured voltage to drop to 1.00V with a steady 0.100A current passing through the circuit?"

I calculated the standard cell potential as 1.05V and the concentration of AgNO3 as 4.46M, but I am completely blanking on what this question is asking.

Explanation / Answer

you calculated AgNO3 molarity = 4.46 M

AgNO3 molarity = 4.46 M

molarity = (W/Mw) x 1000/V

4.46 = (W/170) x 1000/50

W= 0.092g

170g AgNO3 contain ------------------108g Ag+

0.092 g AgNO3 contain-------------?

Ag+ weight = 108 x 0.092 /170

                  = 0.0584 g

equivalent weight (E) = atomic weight/valancy

                                   = 108/1

                                  =108

current (i) =0.1A

time (t) = ?

1F = faraday =96500 C

from faraday's law:

W = E x i x t/96500

0.0584 = 108 x 0.1 x t/96500

t= 522.23 sec

t = 8.70 min

522.23 sec ( or ) 8.70 min time would need.

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