Calculate the cell potential to determine if each of the following balanced redox reactions is spontaneous as written.. Use Table 1 to help you determine the correct half-reactions. Sn(s)+Cu^2+(aq)arrow Sn^2+(aq)+Cu(s) 6 Mg(s)+Pb^2+(aq)arrow Pb(s)+Mg^2+(aq) 7. 2Mn^2+(aq)+8H_(2)O(l)+10Hg^2+(aq)arrow 2MnO_(4)^-(aq)+16H^+(aq)+5Hg_(2)^2+(aq) 8 2SO_(4)^2-(aq)+Co^2+(aq)arrow Co(s)+S_(2)O_(8)^2-(aq) 9 , CHALLENGE Using Table 1 write the equation and determine the cell voltage (E^O) for the following cell.Is the reaction spontaneous?

To determine if each redox reaction is spontaneous, we need to calculate the cell potential (E° cell) using the standard reduction potentials from Table 1. A reaction is spontaneous if the cell potential is positive.<br /><br />1. $Sn(s) + Cu^{2+}(aq) \rightarrow Sn^{2+}(aq) + Cu(s)$<br /><br /> Half-reactions:<br /> $Sn(s) \rightarrow Sn^{2+}(aq) + 2e^-$ (E° = -0.14 V)<br /> $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ (E° = +0.34 V)<br /><br /> Cell potential:<br /> $E°_{cell} = E°_{cathode} - E°_{anode} = 0.34 - (-0.14) = 0.48 V$<br /><br /> Since $E°_{cell}$ is positive, the reaction is spontaneous.<br /><br />2. $Mg(s) + Pb^{2+}(aq) \rightarrow Pb(s) + Mg^{2+}(aq)$<br /><br /> Half-reactions:<br /> $Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-$ (E° = -2.37 V)<br /> $Pb^{2+}(aq) + 2e^- \rightarrow Pb(s)$ (E° = -0.13 V)<br /><br /> Cell potential:<br /> $E°_{cell} = E°_{cathode} - E°_{anode} = 0.13 - (-2.37) = 2.50 V$<br /><br /> Since $E°_{cell}$ is positive, the reaction is spontaneous.<br /><br />3. $2Mn^{2+}(aq) + 8H_{2}O(l) + 10Hg^{2+}(aq) \rightarrow 2MnO_{4}^{-}(aq) + 16H^{+}(aq) + 5Hg_{2}^{2+}(aq)$<br /><br /> Half-reactions:<br /> $Mn^{2+}(aq) \rightarrow MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^-$ (E° = +1.51 V)<br /> $Hg^{2+}(aq) + 2e^- \rightarrow Hg(s)$ (E° = +0.85 V)<br /><br /> Cell potential:<br /> $E°_{cell} = E°_{cathode} - E°_{anode} = 1.51 - 0.85 = 0.66 V$<br /><br /> Since $E°_{cell}$ is positive, the reaction is spontaneous.<br /><br />4. $2SO_{4}^{2-}(aq) + Co^{2+}(aq) \rightarrow Co(s) + S_{2}O_{8}^{2-}(aq)$<br /><br /> Half-reactions:<br /> $Co^{2+}(aq) \rightarrow Co(s) + 2e^-$ (E° = +0.28 V)<br /> $2SO_{4}^{2-}(aq) \rightarrow S_{2}O_{8}^{2-}(aq) + 2e^-$ (E° = +0.80 V)<br /><br /> Cell potential:<br /> $E°_{cell} = E°_{cathode} - E°_{anode} = 0.80 - 0.28 = 0.52 V$<br /><br /> Since $E°_{cell}$ is positive, the reaction is spontaneous.<br /><br />For the challenge question, please provide the specific cell reaction and I will calculate the cell potential for you.