Skip to main content
Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 57b

The metal cadmium tends to form Cd2+ ions. The following observations are made: (i) When a strip of zinc metal is placed in CdCl2(aq), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in Ni(NO3)(aq), nickel metal is deposited on the strip. (b) Which elements more closely define the position of cadmium in the activity series?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
916
views
Was this helpful?

Video transcript

Welcome back everyone When Anthony a metal undergoes oxidation, it commonly forms lengthen iem three plus ions. The ease of oxidation of Anthony um was studied and the following data were gathered in statement one zinc metal deposited on the strip when a strip of land titanium metal was immersed in an aqueous solution of zinc to nitrate. In statement to lengthen E a metal was deposited on the strip when a strip of sodium metal was in an aqueous solution of lengthen iem three nitrate. Based on this information determine which elements best to find Anthony ums position in the activity series. So we want to recall our activity series in our textbooks in which our most reactive metals or the metals that are more easily oxidized are listed towards the top and our least reactive metals. And sorry, this should say most reactive metals, least reactive metals are listed towards the bottom of our activity series in our textbooks. Now going back to the prompt, looking at statement one, if we know that when Lentini a metal undergoes oxidation we form lengthen er three plus ions. And we're told that zinc metal deposits on the strip when Anthony a metal is immersed in a quick solution, this tells us that length any um is oxidized by zinc ions. And so therefore lengthen iem is going to be more reactive than zinc. Whereas in statement to we're told that length ania metal is deposited on the strip when a strip of sodium metal is an Aquarius solution of length any um three nitrate. And so in this case sodium is oxidized by length any um ions and therefore sodium is more reactive. Then lengthen ium And so we can imagine our activity series where sodium would be at the top in our textbooks, zinc would be in the bottom since we're comparing the distance between zinc and sodium since they're the metals mentioned in our prompt and we would find the atoms between or the elements between zinc and magnesium. Sorry, zinc and sodium would be magnesium underneath sodium, aluminum and then we have manganese above zinc. And this means that these three elements would best define the position of length any um in the activity series. Since we understand that sodium is going to be the most reactive aka the most to undergo oxidation, whereas sink being the least reactive since lengthen iem is more reactive than zinc as we defined from our statement one in the prompt and so our final answer highlighted in yellow corresponds to choice C. As the correct choice to complete this example. So I hope this made sense. And let us know if you have any questions