If the solution of the CuSO4 in which copper rod is immersed is dilut…

\(Cu^{2+} + 2e^− \rightarrow Cu\)

\(E_{cell} = E^o_{cell} − \frac{0.0591}{2} log\frac{1}{[Cu^{2+}]}\)

The new concentration of \(Cu^{2+}\) is \(\frac{1}{10}\) of \(Cu^{2+}\)

\(E^1_{cell} = E^o_{cell} − \frac{0.0591}{2}log\frac{10}{[Cu^{2+}]}\)

or, \(E^1_{cell} = E^o_{cell} − \frac{0.0591}{2}log\frac{1}{[Cu^{2+}]} − \frac{0.0591}{2}log 10\)

or, \(\Delta E_{cell} = −\frac{0.0591}{2}log10\)

or, \(\Delta E_{cell} = −\frac{0.059}{2}\)

Thus the electrode potential decreases by \(\frac{0.059 V}{2} = 0.03V\)