1. Identify the half-reactions

Given standard reduction potentials:

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

The reaction is:

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

So:

• $Fe^{3+}$ is reduced to $Fe^{2+}$
• $Sn$ is oxidized to $Sn^{2+}$

2. Write cathode and anode reactions

Cathode (reduction):

$Fe^{3+} + e^- \rightarrow Fe^{2+} \qquad E^\circ = +0.77\,\text{V}$

Anode (oxidation):

The given reduction potential is:

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

Therefore, for oxidation:

$Sn \rightarrow Sn^{2+} + 2e^-$

its oxidation potential is:

$E^\circ_{\text{ox}} = +0.14\,\text{V}$

3. Calculate standard cell potential

Using

$E^\circ_{cell} = E^\circ_{cathode} + E^\circ_{anode(ox)}$

$E^\circ_{cell} = 0.77 + 0.14 = 0.91\,\text{V}$

Alternatively,

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode(red)}$

$E^\circ_{cell} = 0.77 - (-0.14) = 0.91\,\text{V}$

4. Check options

The correct option is:

$\boxed{0.91\,\text{V}}$

So, Option C is correct.