Anode Cate electrolyte Voltage (V)? electrolytelObserved Anode Observations At t

Question

Anode Cate electrolyte Voltage (V)? electrolytelObserved Anode Observations At the anode side, every zinc atom loses two electrons which can form aqueous Zn2; At the cathode side, there will release hydrogen. ZnZn(NO) Hydrogen HC0.76 At side, every zinc atom two electrons which can form ZnH | Zn(NO3)2rt | CuH | Cu(NO),H | 1.10? | aqueous Zn2+. At the cathode side, reduction occurs. The Cu2+ in solution to r, and the solid cooper will plate out on form solid copper the cathode.tf At e side, every zinc atom loses two e can form ZnZn(NOAAgNO56aqueous Zn2; At the cathode side, reduction occurs. The Agt in solution to form solid silver, and the solid silver will plate out on the cathode. At hydronium ion into solution; At the cathode side, reduction occurs. The Cu in solution to form solid copper, and the solid cooper will plate out on the cathode. gen atomm onc Cu Cu(NO)0.34 At gen atom one can forn HydrogenHAg't AgEl | AgNO):1 | 0.80 | hydronunn ion into solution; At the cathode side, reduction occurs. The Ag. in solution to form solid silver, and the solid silver will plate out on the cathode. At side, every cooper atom loses two el s w Cu CuNOAgAgNO046aqueous Cu; At the cathode side, reduction occurs. The Ag in solution to form solid silver, and the solid silver will plate out on the cathode.

Explanation / Answer

From the given data

1. net ionic equations,

Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g)

Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)

Zn(s) + 2Ag+(aq) --> Zn2+(aq) + 2Ag(s)

H2(g) + Cu2+(aq) ---> 2H+(aq) + Cu(s)

H2(g) + 2Ag+(s) ---> 2H+(aq) + 2Ag(s)

Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s)

2. cell notations,

Zn(s) | Zn2+(aq) || HCl(aq) | H2(g)

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)

H2(g) | HCl(aq) || Cu2+(aq) || Cu(s)

H2(g) | HCl(aq) || Ag+(s) | Ag(s)

Cu(s) | Cu2+ || Ag+(aq) | Ag(s)

3. Electric potentials,

Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g): Eo = 0 - (-0.763) = 0.763 V

Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s) : Eo = 0.337 - (-0.763) = 1.100 V

Zn(s) + 2Ag+(aq) --> Zn2+(aq) + 2Ag(s) : Eo = 0.80 - (-0.763) = 1.563 V

H2(g) + Cu2+(aq) ---> 2H+(aq) + Cu(s) : Eo = 0.337 - 0 = 0.337 V

H2(g) + 2Ag+(s) ---> 2H+(aq) + 2Ag(s) : Eo = 0.80 - 0 = 0.80 V

Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s) : Eo = 0.80 - 0.337 = 0.463 V

4. The observed value matches well with the calculate electric potential values.

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