For the cell,Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 , | vedclass

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Question

(B) The cell reaction is: $Mn_{(s)} + Sn^{2+}_{(aq)} \rightarrow Mn^{2+}_{(aq)} + Sn_{(s)}$.Standard cell potential $E^o_{cell} = E^o_{cathode} - E^o_

Vedclass · Accepted Answer

$-194.93$

Vedclass · Answer

(B) The cell reaction is: $Mn_{(s)} + Sn^{2+}_{(aq)} ightarrow Mn^{2+}_{(aq)} + Sn_{(s)}$.Standard cell potential $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{Sn^{2+}/Sn} - E^o_{Mn^{2+}/Mn} = -0.14 - (-1.18) = 1.04 \, V$.The Nernst equation at $298 \, K$ is: $E_{cell} = E^o_{cell} - \frac{0.06}{n} \log \frac{[Mn^{2+}]}{[Sn^{2+}]}$.Here $n = 2$,$[Mn^{2+}] = 0.4$,and $[Sn^{2+}] = 0.04$.$E_{cell} = 1.04 - \frac{0.06}{2} \log \frac{0.4}{0.04} = 1.04 - 0.03 \log(10) = 1.04 - 0.03 = 1.01 \, V$.Free energy change $\Delta G = -nFE_{cell} = -2 	imes 96500 	imes 1.01 \, J = -194930 \, J = -194.93 \, kJ$.

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

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