Solutions

To calculate the electrode potential of zinc when a zinc rod is dipped in a 0.01 M solution of zinc sulfate at 298 K, we use the Nernst equation:

E = E^\circ - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{1}

Given data:

• Standard electrode potential: V
• Concentration of : M
• Temperature: K
• Number of electrons transferred:

Step-by-step calculation:

E = -0.76 - \frac{0.0591}{2} \log (0.01)

Since , we have:

\log (0.01) = \log (10^{-2}) = -2

E = -0.76 - \frac{0.0591}{2} \times (-2)

E = -0.76 + \frac{0.0591 \times 2}{2}

E = -0.76 + 0.0591

E = -0.7009 \, V

Final Answer:

The electrode potential of zinc in this solution is V (approximately).