Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+] = 0.778 M and [Fe2+] = 0.0190 M...
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+]=0.891 M and [Fe2+]=0.0120 M. Use the standard reduction potentials in this table. Cr(s)+Fe2+(aq)↽−−⇀ Cr2+(aq)+Fe(s) E= V
Calculate the cell potential for the reaction as written at 25.00°C, given that [Cr + = 0.827 M and [Ni2+]= 0.0190 M. Use the standard reduction potentials in this table. Cr(s)+Ni2 (aq) Cr2(aq ) + Ni(s) E=
Calculate the cell potential for the reaction as written at 25.00 °C, given that (Cr2+] = 0.773 M and (Ni2+] = 0.0140 M. Use the standard reduction potentials in this table. Cr(s) + Ni2+ (aq) + Cr2+ (aq) + Ni(s) E = 7075
Calculate the cell potential for the reaction as written at 25.00C, given that [Cr2+]=0.785M and [Ni2+]=0.0190M. Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+] = 0.785 M and (Ni2+1=0.0190 M. Use the standard reduction potentials in this table. Cr(s) + Ni2+ (aq) = Cr+ (aq) + Ni(s)
Calculate the cell potential for the reaction as written at 25.00 °C , given that [Cr2+]=0.835 M and [Sn2+]=0.0160 M . Use the standard reduction potentials in this table. Cr(s)+Sn2+(aq)↽−−⇀ Cr2+(aq)+Sn(s) Cr ( s ) + Sn 2 + ( aq ) ↽ − − ⇀ Cr 2 + ( aq ) + Sn ( s )
Calculate the cell potential for the reaction as written at 25.00 °C, given that Cr +1 -0.801 M and [Fe²+] = 0,0120 M. Use the standard reduction potentials in this table. Cr(s) + Fe2+ (aq) = Cr+ (aq) + Fe(s)
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Mg2 Use the standard reduction potentials in this table 0801 M and [Fe+] 0.0120 M. Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Zn2+] = 0.777 M and (Ni2+1 = 0.0190 M. Use the standard reduction potentials in this table. Zn(s) + Ni2+ (aq) = Zn2+ (aq) + Ni(s) V E=
Calculate the cell potential for the reaction as written at 25.00 °C, given that (Zn2+] = 0.885 M and [Ni2+] = 0.0190 M. Use the standard reduction potentials in this table. Zn(s) + Ni2+ (aq) = Zn2+ (aq) + Ni(s) Ev
Calculate the cell potential for the reaction as written at 25.00°C, given that [Zn2+] = 0.758 M and [Fe2+] = 0.0140 M. Use the standard reduction potentials in this table. Zn(s) + Fe2+ (aq) = Zn²+ (aq) + Fe(s) Zn2+(aq) + 2e- → Zn(s) -0.76 Fe2+(aq) + 2e- > Fe(s) -0.44