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Electrochemistry

Question
CBSEENCH12011308

Given, 
(i) Cu2+ + 2e- → Cu,     Eo = 0.337 V

(ii) Cu2+ +e- → Cu+, Eo = 0.153

Electrode potential, Eo for the reaction,
Cu +e- →Cu, will be 

  • 0.52 V

  • 0.90 V

  • 0.30 V

  • 0.38 V

Solution

A.

0.52 V

Gibb's free energy is an additive property.

ΔGo = -nFEo
For reaction, Cu2+ +2e- → Cu;
ΔGo = 2 x F x 0.337  ... (i)
For reaction, Cu+ → Cu2+ +e-;
ΔGo = +1 xF x 0.153
Adding Eqs. (i) and (ii), we get
Cu+ + e- → Cu; ΔGo  = - -0.521 F
 ΔGo   = - nFEo
-0.521 F = -nFEo
Eo = 0.52 V