✔ 最佳答案
16. A is correct. Zn is oxidized to give Zn^2+.
B is correct. Ag^+ is reduced to give Ag coated on the cathode.
C is incorrect. Cl^- in NaCl may react with Ag^+ to form insoluble AgCl, which lowers the electrical conductivity of the cell.
D is correct. Anions carrying -ve charge move towards to ZnNO3 solution to balance the excess amount of +ve charge of Zn^2+ produced.
17. For option A and B, if you don't know which one is correct, just compare these two options. Suppose A is correct, nickel is oxidized so nickel is anode. Then cadmium is cathode, where reduction occurs, then B is also correct. But there could not be two correct answers in MCQ. So they contradict.
A and B are incorrect. Of course in fact, cadmium is more reactive than nickel, so cadmium loses electrons more readily to give oxidation, while nickel ions should be reduced.
C is correct. As mentioned before, cadmium is oxidized to give cadmium ions, so cadmium ions concentration should increase.
D is incorrect. Displacement reaction occurs as cadmium is more reactive than nickel.
18. (1) is correct.
2MnO2 + H2 ---> Mn2O3 + H2O
MnO2 is reduced by H2 formed from zinc-carbon cell. In this case, MnO2 is an oxidizing agent.
(2) is incorrect, the O.N. of Mn decreases from +4 to +3 in the reaction.
(3) is correct, please refer to (1).
Ans: B
19. In half cell A, solution contains Na^+, Br^-.
In half cell B, solution contains K^+, Cr2O7^2-.
Obviously, Cr2O7^2- acts as oxidizing agent in B and Br^- acts as reducing agent. Na^+, K^+ are neither oxidizing agent nor reducing agent because they are too unreactive.
So Cr2O7^2- is reduced, Br^- is oxidized.
Ans: B
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