$$\bbox[ 5px, border: 2px solid red] {\text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \text{Q}_c}$$
\(\text{E}^{\circ}_{\text{sel}} = \text{E}^{\circ}_{\text{zat yang mengalami reduksi}}\:-\:\text{E}^{\circ}_{\text{zat yang mengalami oksidasi}}\)
\(\ce{Al}_{(s)}\: |\: \ce{Al}^{3+}_{\text{(aq)}} (0,36 \text{ M}) \:|| \:\ce{Sn}^{4+}_{\text{(aq)}} (0,086 \text{ M}), \ce{Sn}^{2+}_{\text{(aq)}} (0,54 \text{ M}) \:| \:\ce{Pt}\)
Diketahui \(\text{E}^{\circ}\: \ce{Al}^{3+}\:|\:\ce{Al}_{\text{(s)}} = -1,66 \text{ volt}\) dan \(\text{E}^{\circ}\: \ce{Sn}^{4+}\:|\:\ce{Sn}^{2+} = +0,154\text{ volt}\)
Reaksi oksidasi: \(\ce{Al}_{\text{(s)}} \rightarrow \ce{Al}^{3+}_{\text{(aq)}} + 3e\)
Reaksi reduksi: \(\ce{Sn}^{4+}_{\text{(aq)}} + 2e \rightarrow \ce{Sn}^{2+}_{\text{(aq)}}\)
Selanjutkan samakan jumlah elektron pada kedua reaksi di atas.
Reaksi oksidasi: \(2\ce{Al}_{\text{(s)}} \rightarrow 2\ce{Al}^{3+}_{\text{(aq)}} + 6e\)
Reaksi reduksi: \(3\ce{Sn}^{4+}_{\text{(aq)}} + 6e \rightarrow 3\ce{Sn}^{2+}_{\text{(aq)}}\)
Nilai \(\color{red} n = 6\)
Reaksi sel:
\(2\ce{Al}_{(s)} + 3\ce{Sn}^{4+}_{\text{(aq)}} \rightarrow 2\ce{Al}^{3+}_{\text{(aq)}} + 3\ce{Sn}^{2+}_{\text{(aq)}}\)
\(\text{E}^{\circ}_{\text{sel}} = \text{E}^{\circ}_{\text{reduksi}}\:-\:\text{E}^{\circ}_{\text{oksidasi}}\)
\(\text{E}^{\circ}_{\text{sel}} = +0,154\:-\:(-1,66) = +1,814 \text{ volt}\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \text{Q}_c\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \dfrac{[\ce{Al}^{3+}]^2 \cdot [\ce{Sn}^{2+}]^3}{[\ce{Sn}^{4+}]^3}\)
\(\text{E}_{\text{sel}} = 1,814 \:-\: \dfrac{0,0592}{6} \log \dfrac{(0,36)^2 \cdot (0,54)^3}{(0,086)^3}\)
\(\text{E}_{\text{sel}} = 1,799 \text{ volt}\)
Ion \(\ce{Cr}^{3+}\) dan \(\ce{K}^{+}\) memiliki konsentrasi berturut-turut 0,25 M dan 0,1 M. Kedua ion itu dirangkai pada susunan sel volta. Diketahui \(\text{E}^{\circ}\: \ce{Cr}^{3+}\:|\: \ce{Cr} = -0,71 \text{ V}\) dan \(\text{E}^{\circ}\: \ce{K}^{+}\:|\: \ce{K} = -2,29 \text{ V}\). Tentukan nilai potensial sel volta pada rangkaian tersebut.
Reaksi oksidasi: \(\ce{K}_{\text{(s)}} \rightarrow \ce{K}^{+}_{\text{(aq)}} + e\)
Reaksi reduksi: \(\ce{Cr}^{3+}_{\text{(aq)}} + 3e \rightarrow \ce{Cr}_{\text{(s)}}\)
Selanjutkan samakan jumlah elektron pada kedua reaksi di atas.
Reaksi oksidasi: \(3\ce{K}_{\text{(s)}} \rightarrow 3\ce{K}^{+}_{\text{(aq)}} + 3e\)
Reaksi reduksi: \(\ce{Cr}^{3+}_{\text{(aq)}} + 3e \rightarrow \ce{Cr}_{\text{(s)}}\)
Nilai \(\color{red} n = 3\)
Reaksi sel:
\(3\ce{K}_{\text{(s)}} + \ce{Cr}^{3+}_{\text{(aq)}} \rightarrow 3\ce{K}^{+}_{\text{(aq)}} + \ce{Cr}_{\text{(s)}}\)
\(\text{E}^{\circ}_{\text{sel}} = \text{E}^{\circ}_{\text{reduksi}}\:-\:\text{E}^{\circ}_{\text{oksidasi}}\)
\(\text{E}^{\circ}_{\text{sel}} = -0,71\:-\:(-2,29) = +1,58 \text{ volt}\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \text{Q}_c\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \dfrac{[\ce{K}^{+}]^3}{[\ce{Cr}^{3+}]^1}\)
\(\text{E}_{\text{sel}} = 1,58 \:-\: \dfrac{0,0592}{3} \log \dfrac{0,1^3 }{0,25}\)
\(\text{E}_{\text{sel}} = 1,58 \:-\:(-0,047)\)
\(\text{E}_{\text{sel}} = 1,627 \text{ volt}\)
Diketahui \(\text{E}^{\circ}\: \ce{Cu}^{2+}\:|\:\ce{Cu} = + 0,34 \text{ volt}\) dan \(\text{E}^{\circ}\: \ce{Zn}^{2+}\:|\:\ce{Zn} = -0,76 \text{ volt}\) dimana pada suhu \(25^{\circ}\text{C}\) konsentrasi \(\ce{Zn}^{2+} = 0,25 \text{ M}\) dan konsentrasi \(\ce{Cu}^{2+} = 0,15 \text{ M}\).
Reaksi oksidasi: \(\ce{Zn}_{\text{(s)}} \rightarrow \ce{Zn}^{2+}_{\text{(aq)}} + 2e\)
Reaksi reduksi: \(\ce{Cu}^{2+}_{\text{(aq)}} + 2e \rightarrow \ce{Cu}_{\text{(s)}}\)
Nilai \(\color{red} n =2\)
Reaksi sel:
\(\ce{Zn}_{\text{(s)}} + \ce{Cu}^{2+}_{\text{(aq)}} \rightarrow \ce{Zn}^{2+}_{\text{(aq)}} + \ce{Cu}_{\text{(s)}}\)
\(\text{E}^{\circ}_{\text{sel}} = \text{E}^{\circ}_{\text{reduksi}}\:-\:\text{E}^{\circ}_{\text{oksidasi}}\)
\(\text{E}^{\circ}_{\text{sel}} = +0,34\:-\:(-0,76) = +1,1 \text{ volt}\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \text{Q}_c\)
\(\color{blue} \text{E}_{\text{sel}} = \text{E}^{\circ}_{\text{sel}} \:-\: \dfrac{0,0592}{n} \log \dfrac{[\ce{Zn}^{2+}]}{[\ce{Cu}^{2+}]}\)
\(\text{E}_{\text{sel}} = 1,1 \:-\: \dfrac{0,0592}{2} \log \dfrac{0,25}{0,15}\)
\(\text{E}_{\text{sel}} = 1,1\:-\:0,00657 \text{ volt}\)
\(\text{E}_{\text{sel}} = 1,09\text{ volt}\)
