What is the electric current, in amperes, if 6.50 g Ag is deposited at the cathode in 2.00 hours...?

Question

... in the electrolysis of AgNO3 (aq)?
a. 0.807
b. 48.5
c. 0.442
d. 3.23
e. 1.61

THank you

Answer 1

A lot of people have trouble with this kind of question. The principles are:

Current times time gives you coulombs, which is a measure of amount of electricity

1 Faraday (or 9.6485 x 10^4 C) is a mole of electrons.

You will also need to consider how many electrons are involved in the reaction (in this case, just one because you are reducing Ag+ to Ag).

So

y amps (same as C s-1) x 7,200 s x (1 mol electrons/9.6485E4 C) x (1mol Ag)/(1 mol electrons) x (at. mass Ag)/(1 mol Ag) = 6.50 g Ag

Satisfy yourself that the conversion steps are honest (thing on top worth the same as the thing onthe bottom) and that they take you in the right direction (units cancel correctly). If both these things are true, you can't go wrong.