Please just clearly answer the question, do not show how to do it. Please do not
ID: 797749 • Letter: P
Question
Please just clearly answer the question, do not show how to do it. Please do not post junk from other sites if you aren't sure it is correct, as I've already looked too. Thank you.
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Ag+(aq) + e- rightarrow Ag(s) E degree = +0.80V Co3+(aq) + e- rightarrow Co2+(aq) E degree = +1.82V Cu+2(aq) + e- rightarrow Cu+(aq) E degree = +0.15V Fe3+(aq) + e- rightarrow Fe+2(aq) E degree = +0.77V 2H+(aq) + 2e- rightarrow H2(g) E degree = +0.00V Cl2(g) + 2e- rightarrow E degree = +1.36V Pb2+(aq) + 2e- rightarrow Pb(s) E degree = -0.14V Sn+2(aq) + 2e- rightarrow Sn(s) E degree = -0.14V Zn2+(aq) + 2e- rightarrow Zn(s) E degree = -0.76V Sn4+(aq) + 2e- rightarrow Sn+2(aq) E degree = +0.13VExplanation / Answer
E0 cell = E0 oxidation +E0 Reduction
Given E0=1.53V
From the given values
when Zn--> Zn+2 is taken as oxidation half reaction of cell Eoxidation=-(-0.76)=+0.76
Fe+3--->Fe+2 is taken as reduction half reaction of cell Ereduction=0.77
Therefore Eocell= 0.77+0.73=1.53v