Answer
a)
$E^{0}_{cell} = 2.23 V$
$E_{cell} = 2.23 V$
$ \Delta G = - 430.390 KJ$
b)
$E^{0}_{cell} = 0.02 V$
$E_{cell} = 0.04 V$
$ \Delta G = - 23.160 KJ$
Work Step by Step
a) $Mg_{(s)} + Sn^{2+}_{(aq)} → Mg^{2+}_{(aq)} + Sn_{(s)}$
Here n= 2, since equation involves 2 electrons for charge balancing.
$E_{cell} = E^{0}_{cell} – (RT \div 2F)ln([Mg^{2+}_{(aq)}] \div [Sn^{2+}_{(aq)}])$
$E^{0}_{cell} = E^{0}_{Sn^{2+}/Sn} - E^{0}_{Mg^{2+}/Mg}$
$E^{0}_{cell} = -0.14-(-2.37) = 2.23 V$
[Mg^{2+}] = 0.045 M , [Sn^{2+}] = 0.035 M
$E_{cell} = 2.23 V – (RT \div 2F)ln(0.045 \div 0.035))$
$E_{cell} = 2.23 V – (2.303\times8.314 J K^{-1}\times 298 \div (2\times96500)log(0.045 \div 0.035))$
$E_{cell} = 2.23 V –0.0296 log(0.045 \div 0.035))$
$E_{cell} = 2.23 V –0.0296 \times 0.1091$
$E_{cell} = 2.23 V –0.0032 = 2.23 V$
$ \Delta G = -n F E_{cell}$
$ \Delta G = -2\times 96500 \times 2.226 = 430390 J =- 430.390 KJ$
b) $3Zn_{(s)} +2Cr^{3+}_{(aq)} + → 2Cr_{(s)} + 3Zn^{2+}_{(aq)} $
Here n= 6, since equation involves 6 electrons for charge balancing.
$E_{cell} = E^{0}_{cell} – (RT \div 6F)ln([Zn^{2+}_{(aq)}]^{3} \div [Cr^{3+}_{(aq)}]^{2})$
$E^{0}_{cell} = E^{0}_{Cr^{3+}/Cr} - E^{0}_{Zn^{2+}/Zn}$
$E^{0}_{cell} = -0.74-(-0.76) = 0.02 V$
[Cr^{3+}] = 0.010 M , [Cr^{3+}] = 0.0085 M
$E_{cell} = 0.02 V – (RT \div 6F)ln([0.0085]^{3} \div [0.010]^{2})$
$E_{cell} = 0.02V – (2.303\times8.314 J K^{-1}\times 298 )\div (6\times96500)log(0.0085 \div 0.010)$
$E_{cell} = 0.02 V –0.0099 log(6.1\times 10^{-7} \div 1\times 10^{-4})$
$E_{cell} = 0.02 V –0.0099 log (0.0061)$
$E_{cell} = 0.02 V –0.0099 \times(- 2.2147)$
$E_{cell} = 0.02 V –0.02 = 0.04 V$
$ \Delta G = -n F E_{cell}$
$ \Delta G = -6\times 96500 \times 0.04 = 23160 J = - 23.160 KJ$
