Question

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

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

Answer 1

Answer: (B)

$Sn(s)+2F{e}^{3+}(aq)\to 2F{e}^{2+}(aq)+S{n}^{2+}(aq)$
${\text{E}}_{\text{cell}}^{\text{o}}={\left({\text{E}}_{\text{reduction}}^{\text{o}}\right)}_{\text{cathode}}-{\left({\text{E}}_{\text{reduction}}^{\text{o}}\right)}_{\text{anode}}$
${\text{E}}_{\text{cell}}^{\text{o}}=0.77-(-0.14)$
$\therefore {\text{E}}_{\text{cell}}^{\text{o}}=-0.91\text{V}$