Given the cell reaction,Zn(s) + Cut2Cu(s)Cu +2the cathode in the cell is:O Pt(aq)Zn(s)(aq)Zn(aq)→ Zn¹2 + Cu(s)+2(aq) The standard (half cell) reduction potentials of two reactions are given below.Li+ (aq) + eF2(g) + 2e¯(aq)From this information, one may conclude that Li(s):Li(s)2Fis an excellent reducing agentis an excellent oxidizing agentis very difficult to oxidizewill not react with F2(g)is very electronegativeEº = - 3.04 VE° +2.87 V

1) Zn(s) + Cu2+(aq) ==> Zn2+(aq) + Cu(s) As you have given two questions , but according to the…

Answered: Zn(s) O Cu(s) the cathode in the cell…

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Zn(s) O Cu(s) the cathode in the cell is: O Cu ¹2 OPt (aq) Zn(s) + Cu+2 Zn +2 + Cu(s) (aq) → Zn+2 (aq) (aq)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Given the cell reaction,
Zn(s) + Cut2
Cu(s)
Cu +2
the cathode in the cell is:
O Pt
(aq)
Zn(s)
(aq)
Zn(aq)
→ Zn¹2 + Cu(s)
+2
(aq)

The standard (half cell) reduction potentials of two reactions are given below.
Li+ (aq) + e
F2(g) + 2e¯
(aq)
From this information, one may conclude that Li(s):
Li(s)
2F
is an excellent reducing agent
is an excellent oxidizing agent
is very difficult to oxidize
will not react with F2(g)
is very electronegative
Eº = - 3.04 V
E° +2.87 V

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