Question

Consider the following $E^{o}$ values
$E_{F e^{3 +} / F e^{2 +}}^{o}=0.77V$
$E_{S n^{2 +} / S n}^{o}=-0.14V$
Under standard conditions the potential for the reaction
$Sn\left(\right.s\left.\right)+2Fe^{3 +}\left(\right.aq\left.\right) \rightarrow 2Fe^{2 +}\left(\right.aq\left.\right)+Sn^{2 +}\left(\right.aq\left.\right)$

• A. 1.68 V
• B. 0.91 V
• C. 0.63 V
• D. 1.46 V

Answer 1

Answer: (B)

$Sn\left(\right.s\left.\right)+2Fe^{3 +}\left(\right.aq\left.\right) \rightarrow 2Fe^{2 +}\left(\right.aq\left.\right)+Sn^{2 +}\left(\right.aq\left.\right)$
$\text{E}_{\text{cell}}^{\text{o}} = \left(\text{E}_{\text{reduction}}^{\text{o}}\right)_{\text{cathode}} - \left(\text{E}_{\text{reduction}}^{\text{o}}\right)_{\text{anode}}$
$\text{E}_{\text{cell}}^{\text{o}} = 0.77 - \left(\right. - 0.14 \left.\right)$
$\therefore \, \text{E}_{\text{cell}}^{\text{o}} = - 0.91 \, \text{V}$