Question
Asked Mar 15, 2019

Two Pt electrodes are placed in an aqueous solution of ZnSO4 at pH 7 and a potential of 2.00v is applied to the electrodes. In the sulfate ion, sulfur is in its highest oxidation state. 

a) What are the possible reduction reactions that can take place in this system? 

(2H+) + (2e-)--> H2           E= -0.41v at pH 7

(Zn^2+) + (2e-)--> Zn       E*= -0.76v 

Can you explain why this is the solution? 

Useful
Information
Half reaction
E (v)
½Cl2 + e-→ Cl-
+1.36 v
0.82 (pH 7.00)
0.80
0.77
0.34
-0.036
-0.41 (pH 700)
0.44
0.76
Ag+ + e-→ Ag(s)
Fe te- Fe
Cut +2e Cu(s)
Fet +3e Fe(s)
Zn2 +2e Zn(s)
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Useful Information Half reaction E (v) ½Cl2 + e-→ Cl- +1.36 v 0.82 (pH 7.00) 0.80 0.77 0.34 -0.036 -0.41 (pH 700) 0.44 0.76 Ag+ + e-→ Ag(s) Fe te- Fe Cut +2e Cu(s) Fet +3e Fe(s) Zn2 +2e Zn(s)

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Expert Answer

Step 1

Here, the half-cell reactions are shown as reduction.

 

That is, increasing oxidizing power potential or decreasing reduction power potential.

 

Step 2

Thus the possible reduct...

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