Calculate the potential of a platinum electrode immersed in a solution of:
a) K_2PtCl_4 \space 0.016M \space y \space KCl \space 0.2450M.
PtCl_4^{2-} + 2e^{-} \leftrightarrows Pt_{(s)} + 4Cl⁻ \space E^o = 0.755V .
E = 0.755V - \frac{0.0592}{2} \times log (\frac{0.2450^2}{0.016}) .
E = 0.774V .
b) Sn(SO_4)_2 \space 0.0650M \space y \space SnSO_4 \space 3.5 \times 10^{-3}M.
Sn^{4+} + 2e^{-} \leftrightarrows Sn^{2+} \space E^o = 0.154V .
E = 0.154V - \frac{0.0592}{2} \times log(\frac{3.5 \times 10^{-3}}{0.065}) = 0.192V.
c) pH = 6.50 \space H_2 \space 1atm .
2H^{+} + 2e^{-} \leftrightarrows H_2 \space E^{o} = 0.000V.
E = \frac{-0.0592}{2} \times log(\frac{1}{(10^{-6.50})^2}) .
E = -0.385V .
d) VOSO_4 \space 0.0255M, \space V_2(SO_4)_3 \space 0.0686M, \space HClO_4 \space 0.100M .
VO^{2+} + 2H^{+} + e^{-} \leftrightarrows V^{3+} + H_2O \space E^o = 0.337V .
E = 0.337V – \frac{0.0592}{1} log(\frac{0.0686 \times 2}{(0.0255)(0.100)^2}) .
E=0.175V .
e) 25.00mL \space SnCl_2 \space 0.0918M, \space 25.00mL FeCl_3 \space 0.1568M .
Sn^{4+} + 2e^{-} \leftrightarrows Sn^{2+} \space E^o=0.154V .
Fe^{3+} + e^{-} \leftrightarrows Fe^{2+} \space E^o=0.771V .
2Fe^{3+} + Sn^{2+} \leftrightarrows 2Fe^{2+} + Sn^{4+} .
25.00mL \times \frac{0.0918mol}{1000mL} = 2.295 \times 10^{-3} mol Sn^{2+} .
25.00mL \times \frac{0.1568mol}{1000mL} = 3.92 \times 10^{-3} mol Fe^{3+} .
2.295 \times 10^{-3} mol Sn^{2+} \times \frac{2molFe^{3+}}{1molSn^{2+}} = 4.59 \times 10^{-3} mol Fe^{3+} .
limiting reagent Fe^{3+} .
3.92 \times 10^{-3} molFe^{3+} \times \frac{1molSn^{4+}}{2molFe^{3+}} = 1.96\times 10^{-3}molSn^{4+} .
2.295 \times 10^{-3} mol Sn^{2+} – 1.96\times 10^{-3}molSn^{2+} = 0.335\times 10^{-3} mol Sn^{2+} .
E = 0.154V – \frac{0.0592}{2} log(\frac{0.335\times 10^{-3} mol Sn^{2+}/{0.05L}}{{1.96\times 10^{-3}molSn^{4+}}/{0.05L}}) .
E = 0.177V .