Answer
a) $E_{cell} = 2.23 V – (RT \div 2F)ln([Mg^{2+}_{(aq)} \div [Sn^{2+}_{(aq)}])$
b) $E_{cell} = 0.61 V – (RT \div 6F)ln([Cr^{3+}_{(aq)}]^{2} \div [Pb^{2+}_{(aq)}]^{3})$
Work Step by Step
Nernst equation for a general reaction,
a A + bB → c C + d D
The above reactions Nernst equation can be written as,
$E_{cell} = E^{0}_{cell} – (RT \div nF)ln([C]^{c}[D]^{d} \div [A]^{a}[B]^{b})$
a) $Mg_{(s)} + Sn^{2+}_{(aq)} → Mg^{2+}_{(aq)} + Sn_{(s)}$
Here n= 2, since equation involves 2 electrons for charge balancing.
$E_{cell} = E^{0}_{cell} – (RT \div 2F)ln([Mg^{2+}_{(aq)}] \div [Sn^{2+}_{(aq)]})$
$E^{0}_{cell} = E^{0}_{Sn^{2+}/Sn} - E^{0}_{Mg^{2+}/Mg}$
$E^{0}_{cell} = -0.14-(-2.37) = 2.23 V$
Final Nernst equation is
$E_{cell} = 2.23 V – (RT \div 2F)ln([Mg^{2+}_{(aq)} \div [Sn^{2+}_{(aq)]})$
b) $2Cr_{(s)} + 3Pb^{2+}_{(aq)} → 2Cr^{3+}_{(aq)} + 3Pb_{(s)}$
Here n= 6, since equation involves 6 electrons for charge balancing.
$E_{cell} = E^{0}_{cell} – (RT \div 6F)ln([Cr^{3+}_{(aq)}]^{2} \div [Pb^{2+}_{(aq)}]^{3})$
$E^{0}_{cell} = E^{0}_{Pb^{2+}/Pb} - E^{0}_{Cr^{3+}/Cr}$
$E^{0}_{cell} = -0.13-(-0.74) = 0.61 V$
Final Nernst equation is
$E_{cell} = 0.61 V – (RT \div 6F)ln([Cr^{3+}_{(aq)}]^{2} \div [Pb^{2+}_{(aq)}]^{3})$
