The last calculated result on the report page is the ratio of moles of silver ob
ID: 534363 • Letter: T
Question
The last calculated result on the report page is the ratio of moles of silver obtained to the moles of copper wire that reacted. Predict the effect that the following experimental errors would have on this ratio: would the result appear too high (H), too low (L), or would there be no effect (N)? Briefly justify your answers. [In this kind of question, you assume you are unaware that the error occurred.] The silver had not been completely dried. _______ The original mass of the copper wire had mistakenly been recorded too low.___________ Some of the solution spattered out the beaker in step 2. __________ Copper metal is used to precipitate silver ions from solution (by reducing them to silver metal). Could copper also cause zinc ions to precipitate in this way? (Explain.) A 2.000 g piece of copper wire was used to precipitate silver ions as silver from a solution. When the copper metal reacts, Cu^2+ ions are formed, which are soluble in water. If 0.750 g of Ag(s) were produced in this process, what should the final mass of the copper wire be?Explanation / Answer
3. In this reaction silver is precipitated as using copper.
If the silver obtained from the reaction is not completely dried. Then we will get more product then the theoretical yield. Then products yield is very high, (H).
For solutions in water one copper can reduce 2 silver ions (The stable oxidation state of copper in water is +2 and silver is +1).
Thus, if silver is not dried properly the reaction yields experimental result appear as too large. (H)
b. If the mass of Cu is low mistakenly recorded as low value, then also experimental result appear as too large (H) in yield. Since the reaction is taking place in molar quantity, the number of moles in the copper wire is more than that of calculated. Also, the final yield is calculated by considering the moles of silver formed to the moles of Cu used. Thus if the recorded mass of Cu is too low. It appears as H
c. At initial time solution contains silver ions. If solution is spilled, silver ions are lost. After the reaction, Ag precipitate is dried and weighed for calculating the yield of the reaction. If some amount of silver ions are lost from the solution, the amount of Ag precipitated will be less from the calculated value. (L)
As per the policy of chegg first question is answered.
2. No. Becouse reduction potential of Zn is less than that of Cu. Then Zn will not get reduced in the presence of Cu
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