For a cell, $$Cu(s)|Cu^{2+}(0.001 M)||Ag^+(0.01 M)|Ag(s)$$, the cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for $$Cu^{2+}|Cu$$ is _____ $$\times 10^{-2}$$ V. Given: $$E^\theta_{Ag^+/Ag} = 0.80$$ V and $$\frac{2.303RT}{F} = 0.06$$ V

We have the cell $$Cu(s)|Cu^{2+}(0.001$$ M $$)||Ag^+(0.01$$ M $$)|Ag(s)$$ with a cell potential of 0.43 V at 298 K. We are given $$E^\theta_{Ag^+/Ag} = 0.80$$ V and $$\frac{2.303RT}{F} = 0.06$$ V. We need to find the magnitude of the standard electrode potential for $$Cu^{2+}/Cu$$, expressed as $$\_\_\_ \times 10^{-2}$$ V.

The cell reaction is: $$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$$

The standard cell potential is:

$$E^\theta_{cell} = E^\theta_{cathode} - E^\theta_{anode} = E^\theta_{Ag^+/Ag} - E^\theta_{Cu^{2+}/Cu} = 0.80 - E^\theta_{Cu^{2+}/Cu}$$

Now we apply the Nernst equation. For this reaction, the number of electrons transferred is $$n = 2$$:

$$E_{cell} = E^\theta_{cell} - \frac{0.06}{n}\log Q$$

The reaction quotient is:

$$Q = \frac{[Cu^{2+}]}{[Ag^+]^2} = \frac{0.001}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10$$

Substituting into the Nernst equation:

$$0.43 = E^\theta_{cell} - \frac{0.06}{2}\log 10$$

$$0.43 = E^\theta_{cell} - 0.03 \times 1$$

$$E^\theta_{cell} = 0.43 + 0.03 = 0.46$$ V

Now we can find $$E^\theta_{Cu^{2+}/Cu}$$:

$$0.46 = 0.80 - E^\theta_{Cu^{2+}/Cu}$$

$$E^\theta_{Cu^{2+}/Cu} = 0.80 - 0.46 = 0.34$$ V

Expressed as $$\_\_\_ \times 10^{-2}$$ V, we get $$0.34 = 34 \times 10^{-2}$$ V.

Hence, the correct answer is 34.