A galvanic cell based on the reaction 3Ag^+(aq) + Al(s) rightarrow 3 Ag(s) + Al^
ID: 519244 • Letter: A
Question
A galvanic cell based on the reaction 3Ag^+(aq) + Al(s) rightarrow 3 Ag(s) + Al^3+(aq) is constructed from two half-cells: one with a silver electrode in 1.0 M AgNO_3 and an aluminum electrode in 10 M Al(NO_3)_2. The two half-cells are connected with a wire and a salt bridge filled with 1.0 M Na_2SO_4. As the reaction proceeds, which electrode (Ag or Al) will decrease in mass? At the end of the reaction, which half-cell (Ag^+/Ag or Al/Al^3+) will contain sulfate ion? Based on the activity series in your textbook (Table 9.6, page 338), which of the following metals will not react with a strong acid? There may be more than one right answer! Mercury Chromium Copper Cadmium What is the value of E degree _cell, in volts, for the galvanic cell produced by connecting the following half-cells? Enter your answer to the thousandths place. Mn^2+(aq) + 2e^- Equilibrium Mn(s) E_red degree = -1.185 V In^3+(aq) + 3e^- Equilibrium In(s) E_red degree = -0.3382 VExplanation / Answer
Solution:- (5) From the activity series, Ag comes after Al it means Ag could be reduced by Al. So, as the reactions proceeds, Al is oxidized to Al3+ and Ag+ is reduced to Ag. In this process, the oxidation of Al takes place at the anode and the Al metal goes from electrode to the solution into the form of Al3+ ions. So, the mass of Al electrode decreases.
(6) in the galvanic cell, the salt bridge is used for the charge balance. since the Al3+ ions goes to the slution from the anode so the anodic solution becomes rich in positive ions. to balance these positive ions, the negative ions that is SO42- (sulfate) ions comes into the solution from the salt bridge.
So, the right answer is, the Al/Al3+ contains the sulfate ions.
(7) In activity series only the metals that are above hydrogen, react with the acids. Cr and Cd are above hydrogen so they could react with acids. Hg and Cu comes below hydrogen so they would not react with acids.
So, the right choices are a) Mercury and c) copper.
(8) E0cell = E0(cathode) - E0(anode)
the electrode for which standard reduction potential is more negative or less positive acts as anode.
If we look at the given standard reduction potentials then it is more negative for first half equation. So, that's anode and second half one is cathode.
E0cell = -0.3382 V - (-1.185 V) = -0.3382 V + 1.185 V
E0cell = +0.8468 V
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