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Practice quiz in answering step by step standard cell potential
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Names : DEGUINIO, Gerald Angelo D. Section : BI3A EXCONDE, Joaquin Adrian M. VILLAFUERTE, Neil Patrick V. A. 2Na(s) + Fe2+^ (aq) ---------> 2Na+(aq) + Fe(s) ANODE : 2Na โ 2Na+^ + 1e-^ E^0 = +2.71 V CATHODE : Fe2+^ + 2e-^ โ Fe E^0 = - 0.44 V 2 (2Na โ 2Na+^ + 1e-^ ) Fe2+^ + 2e-^ โ Fe BALANCED EQ : 4Na + Fe2+^ โ 4 Na+^ + Fe E^0 = +2.27 V (Spontaneous) B. 3SO 42 -^ (aq) + 12H+^ (aq) + 2Al (s) ---> 3SO 2 (g) + 2Al3+(aq) + 6H 2 O ANODE : 2Al โ 2Al3+^ + 3 e-^ E^0 = +1.66 V CATHODE : 3SO 42 -^ + 12H+^ + 2e-^ โ 3SO 2 + 6H 2 O E^0 = +0.17 V 2(2Al โ 2Al3+^ + 3 e-^ ) 3( 3SO 42 -^ + 12H+^ + 2e-^ โ 3SO 2 + 6H 2 O ) BALANCED EQ : 4Al + 6 SO 42 + 36 H+^ โ 4 Al3+^ + 9 SO 2 + 18 H 2 O E^0 = +1.83 V
C. 2S (s) + 4OH-^ (aq) --------> O 2 (g) + 2S-^2 (aq) + 2H 2 O (basic ) ANODE : 4OH-^ โ O 2 + 2H 2 O + 4e-^ E^0 = - 0.40 V CATHODE : 2S + 2e-^ โ 2S-^2 E^0 = - 0.4 5 V 4OH-^ โ O 2 + 2H 2 O + 4e- 2(2S + 2e-^ โ 2S-^2 ) BALANCED EQ : 4 S + 4OH-^ โ 4 S-^2 + O 2 + 2H 2 O E^0 = - 0.85 V (Non-spontaneous) D. Chromium (II) ions and tin (IV) ions to produces chromium (III) ions and tin(II) ions ANODE : Cr2+^ โ Cr+3^ + 1e-^ E^0 = +0. 50 V CATHODE : Sn+4^ + 2e-^ โ Sn+^2 E^0 = +0. 15 V 2(Cr2+^ โ Cr+3^ + 1e-^ ) Sn+4^ + 2e-^ โ Sn+^2 BALANCED EQ : 2 Cr2+^ + Sn+4^ โ 2 Cr+3^ + Sn+^2 E^0 = +0. 65 V (Spontaneous) E. Manganese (II) ions and hydrogen peroxide to produce solid manganese dioxide (MnO 2 ). ANODE : Mn2+^ + 2H 2 O โ MnO 2 + 4H+^ + 2eโ E^0 = - 1. 21 V CATHODE : H 2 O 2 + 2H+^ + 2eโ โ H 2 O E^0 = +1.78 V Mn2+^ + 2H 2 O โ MnO 2 + 4H+^ + 2eโ H 2 O 2 + 2H+^ + 2eโ โ H 2 O BALANCED EQ : Mn2+^ + H 2 O + H 2 O 2 โ Mn2+^ + 2H+^ E^0 = + 0. 57 V