Consider the following $$E^\circ$$ values: $$E^\circ_{Fe^{3+}/Fe^{2+}} = 0.77$$ V; $$E^\circ_{Sn^{2+}/Sn} = -0.14$$ V. Under standard conditions the potential for the reaction $$Sn(s) + 2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq) + Sn^{2+}(aq)$$ is

Given:

$$E^\circ_{Fe^{3+}/Fe^{2+}} = 0.77\;V$$
$$E^\circ_{Sn^{2+}/Sn} = -0.14\;V$$

Reaction:

$$Sn(s) + 2Fe^{3+}(aq)\rightarrow2Fe^{2+}(aq) + Sn^{2+}(aq)$$

Oxidation half-reaction:
$$Sn(s) \rightarrow Sn^{2+} + 2e^-$$

Reduction half-reaction:

$$2Fe^{3+} + 2e^-\rightarrow2Fe^{2+}$$

Cathode (reduction):

$$Fe^{3+} + e^- \rightarrow Fe^{2+}$$

$$E^\circ_{\text{cathode}} = 0.77\;V$$

Anode (oxidation):
$$Sn \rightarrow Sn^{2+} + 2e^-$$

Given reduction potential:

$$Sn^{2+} + 2e^- \rightarrow Sn$$
$$E^\circ = -0.14\;V$$

Therefore oxidation potential:

$$E^\circ_{\text{oxidation}} = +0.14\;V$$

Cell potential:

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}+E^\circ_{\text{oxidation}}$$ 
$$= 0.77 + 0.14$$ $$= 0.91\;V$$

Therefore, $${E^\circ_{\text{cell}} = 0.91\;V}$$