Question

8.3 Standard Reduction Po Also, since the two half-reactions invo anust be multiplied by integers as foilows lve different numhers of electrons 3(Mg--> Mg2+ + 2e-) %" (cathode)--1.66 V 2.37 V 2A13+ (aq) + 3Mg(s)-→ 2Al(s) + 3Mgat (aq) go (anode) l(cathode) (anode -1.66 V + 2.37 V 0.71 b. Halfr I-reaction (2) must be reversed (it is the anode), and both half-reactions must e multiplied by integers to make the number of electrons equal: (cathode) 1.51v + H2O一一→ Cl04_ + 2H+ + 2e-) -go (anode) =-1.19 V 2(MnO4-+ 5e_ + 8H+-→ Mn2+ + 4H2O) ー з 2Mno,-(aq) + 6H+ (aq) + 5C103-(aq)-→ 酱eell %" (cathode)-%" (anode) = 1.51 V-1.19 V 0.32 V See Exercises 18.39 and 18.40 2Mn2 (aą) 3H,00) 5C1Os Caq) Line Notation We now will introduce a handy line notation used to describe electrochemical cells. In this notation the anode components are listed on the left and the cathode components are listed on the right, separated by double vertical lines (indicating the salt bridge or r the line notation for the cell described in Example 18.3(a) is

how is cell potential become 1.51 - 1.19?? for question b.
it was originally 1.51 / 1.19 both positive.
then reversed second reaction (anode)
shouldn't it be 1.51 - (-1.19) so 2.7..?