2. Procedural question from Part B: You will need to mix 20 drops of 0.10 M copp
ID: 554670 • Letter: 2
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2. Procedural question from Part B: You will need to mix 20 drops of 0.10 M copper(Il) sulfate and 20 drops of 7M NH, Why do the drops need to be the same size? a) Why do the drops need to be the same size? b) After mixing 20 drops of each reagent, what will be the new concentration of each? 40 drp 13.SM 3. Procedural question from Part C: In step 2, you are asked to prepare a new voltaic cell. After following instructions, you see a very faint white precipitate form. What is the precipitate? 4. Write the half-cell reactions associated with the following oxidation-reduction reaction: (be sure to include ion charges and phases associated with each species) Ni(s) + Hg22+(aq) 2Hg(l) + Ni2+ (aq) Anode half-cell reaction: Cathode half-cell reaction 175Explanation / Answer
2a) The dilution equation can be applied only when all the drops have the same size. This is because we assume that when the drops are of the same size, the volume of each drop will be the same. Consequently, it is imperative that all the drops have the same size.
3) NH3 dissolves in water to form aqueous ammonia, also represented as NH4OH. The reaction for the formation is
NH3 (aq) + H2O (l) -------> NH4OH (aq)
CuSO4 reacts with aqueous ammonia to form a precipitate of copper hydroxide, Cu(OH)2. This precipitate is pale white in color. The precipitate dissolves in excess ammonia to form a deep blue copper-amine complex.
CuSO4 (aq) + 2 NH4OH (aq) ---------> Cu(OH)2 (s) + (NH4)2SO4 (aq)
4) It is clear that Ni is oxidized to Ni2+ while Hg22+ is reduced to Hg. Oxidation takes place at the anode and reduction occurs at the cathode. Therefore, the half reactions are
Anode half-cell reaction: Ni (s) --------> Ni2+ (aq) + 2 e-
Cathode half-cell reaction: Hg22+ (aq) + 2 e- ---------> 2 Hg (l)
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