In basic solution Se2- and SO32- ions react spontaneously: 2Se_(aq)^2-+2SO_3(…

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In basic solution Se2- and SO32- ions react spontaneously:
2Se(aq)2βˆ’+2SO3(aq)2βˆ’+3H2O(l)β†’2Se(s)+6OH(aq)βˆ’+S2O(aq)2βˆ’,Ecello=0.35V2Se_{(aq)}^{2-}+2SO_{3(aq)}^{2-}+3H_2O_{(l)} \to 2Se_{(s)}+6OH_{(aq)}^-+S_2O^{2-}_{(aq)}, E^o_{cell}=0.35V
  1. Write balanced half reactions for this process
  2. If Esulfite is -0.57 V, calculate Eselenium.
More Reduction Potentials Questions:
(Duplicated)
Use the standard reduction potentials below to answer the following question: which species listed is the strongest oxidizing agent?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}
Electrochemistry: Reduction potentials in a Galvanic Cell
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}
Use the standard reduction potentials below to answer the following question: which metal could be used as an anode for an aluminum pipe?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}