Q13 of 27 Page 1

Zinc rod is dipped in 0.01 M solution of zinc sulphate when temperature is 298 K. Calculate the electrode potential of zinc. (Given: E° (Zn2+/Zn) = – 0.76 V; log 10 = 1)

Given, E° (Zn2+/Zn) = – 0.76 V


Zn2+ + 2e-→ Zn


Concentration of ZnSO4 = 0.01 M


Temperature = 298 K


According to Nernst Equation,


Ecell = E° -


= - 0.76 -


= -0.76 – 0.059


= - 0.819 V


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