A galvanic cell was constructs employing the following half-reactions: Cu^2+ + 2

Question

A galvanic cell was constructs employing the following half-reactions: Cu^2+ + 2e^- rightarrow Cu E_0 = +0.34 V Sn^2+ + 2e^- rightarrow E_0 = - 0.14 V Calculate the cell potential for the spontaneous process. Consider the following half-reactions and their standard reduction potentials: Ce^4+ + e^- rightarrow Ce^3+ +1.61 V Hg^2+ + 2 e^- rightarrow Hg_(l) + 0.79 V Sn^2+ + 2 e^- rightarrow Sn - 0.14 V Al^3+ + 3e^- rightarrow Al - 1.66 V (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will Sn reduce Al^3+ to Al? (f) List the ions above that can be oxidized by Sn^2+.

Explanation / Answer

4)

from data table:

Eo(Sn2+/Sn(s)) = -0.14 V

Eo(Cu2+/Cu(s)) = 0.34 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Cu2+/Cu(s))

anode is (Sn2+/Sn(s))

The chemical reaction taking place is

Cu2+(aq) + Sn(s) --> Cu(s) + Sn2+(aq)

Eocell = Eocathode - Eoanode

= (0.34) - (-0.14)

= 0.48 V

Answer: 0.48 V

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