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

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(B) The cell reaction is: $Mn_{(s)} + Sn_{(aq)}^{2+} \rightarrow Mn_{(aq)}^{2+} + Sn_{(s)}$.The standard cell potential is $E_{cell}^o = E_{cathode}^o

Vedclass · Accepted Answer

$-194.93$

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

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

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