Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction: Ag+(aq) + e- ¡ Ag(s) What mass of silver would plate onto the cathode if a current of 6.8 A flowed through the cell for 72 min?

Video transcript

Hey everyone in this example, we're told that at the cathode of an electrolytic cell, cadmium can be electro plated with the following half reaction. We need to calculate the mass of cadmium that is plated onto the cathode when 3.2 ampules of a current goes through the cell for minutes. So what we're going to begin with is the time given from the prompt, 56 minutes And we want to go ahead and convert from minutes to seconds. So we're going to multiply by the conversion factor where we recall that in one minute. We have 60 seconds. So now we're able to cancel out our units of minutes. And now we're going to multiply by the next conversion factor where we should recall that we have one ampule equal to a column per second. And so according to the prompt, we have 3.2 ampules which we can understand as 3.2 Columns per second, meaning that for one second we have 3.2 columns as our next conversion factor. So now we can cancel out seconds. And now we're focusing on converting from columns to moles. And so we should recall our Faraday's constant Which says that we have a value of 96,485 columns Per mole of electron. And so we would plug that in as 96,485 columns for one mole of an electron. And so now we're able to cancel out our units of columns and we want to refer to our equation here where we see that for our ratio of cadmium two plus carry on to our electron electrons that are transferred. We have a 1-2 ratio here. So we can multiply this as a conversion factor to go from one mole of our cadmium two plus caddy on, which will correspond to two moles of electrons that were transferred. And this came directly from our equation here. So now we're able to cancel out our units of moles of electrons and moles of cadmium here And sorry, that should be flipped around. So in our denominator we want to plug in two moles of electrons that are transferred for one mole of our cadmium two plus catalon. And so now it makes sense that we can cancel out moles of electrons. And so for the final part of this question we need to calculate massive cadmium. So we're going to multiply by our last conversion factor. Sorry about that. So our last conversion factor we would recall. And just to make more room, I'm gonna slide this all over. So we're going to recall that we should go from moles of our adam cadmium two g of our adam cadmium to answer this question. And so we would recall from our periodic tables that for one mole of cadmium we have a molar mass of 112.411 g of cadmium. And so now we're able to cancel out moles of cadmium with moles of cadmium. And we're left with our final unit, which is grams of cadmium. And so now we're going to get for our final answer a value equal to 6.26 g of cadmium. And so this value here would be our final answer to complete this example as our mass of cadmium that is placed onto our cathode when 3.2 ampules of a current goes through the cell for 56 minutes, so it's highlighted in yellow. Here is our final answer. I hope that everything I explained was clear. If you have any questions, just leave them down below. Otherwise I will see everyone in the next practice video.