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]
Answer & explanation
Correct answer: option 2
The electrode reaction is given as
Zn+2 + 2e- → Zn
Using Nernst Equation
EZn2+/Zn = E°Zn2+/Zn - \(\frac{0.0591}{2}\)log\(\frac{1}{[Zn^{2+}]}\)
[Zn2+] = \(\frac{95}{100}\)x0.1 = 0.095
EZn2+/Zn = -0.76 - 0.0295log\(\frac{1}{[0.095]}\)
EZn2+/Zn = -0.76 - 0.0295[log1000 - log95]
EZn2+/Zn = -0.76 - 0.0295[3 - 1.9777]
EZn2+/Zn = -0.76 - 0.03016 = -0.79 V