A zinc rod is dipped in 0.1 M solution of ZnSO₄. The salt is 95% dissociated at this dilution at 298 K. What is the electrode potential?
[E°Zn²⁺/Zn = – 0.76 V] [log95 = 1.9777]

-0.79 V

The electrode reaction is given as
Zn⁺² + 2e^- → Zn 
Using Nernst Equation

E_Zn²⁺/Zn = E°_Zn²⁺/Zn - \(\frac{0.0591}{2}\)log\(\frac{1}{[Zn^{2+}]}\) 

[Zn²⁺] = \(\frac{95}{100}\)x0.1 = 0.095

E_Zn²⁺/Zn = -0.76 - 0.0295log\(\frac{1}{[0.095]}\) 

E_Zn²⁺/Zn = -0.76 - 0.0295[log1000 - log95]

E_Zn²⁺/Zn = -0.76 - 0.0295[3 - 1.9777]

E_Zn²⁺/Zn = -0.76 - 0.03016 = -0.79 V