Assume that the reduction potential for your most active metal is 0.0V. Using yo

Question

Assume that the reduction potential for your most active metal is 0.0V. Using your observed voltages, determine the reduction potentials for the other oxidized species investigated.

Could someone explain number 7 for me. If you could do 8 and 9 would also be very helpful thank you.

Cell Cathode Voltage Anode Zn Cu Zn cu 0,9 Zn Fe Zn Fe 0, 574 Ni 1.33 Zn Ni Zn 12/1- Zn. Cu Fe Fe cu 0. Cu Ni Cu 12/l Fe Ni Fe 0.809 Fe 12/I Ni 12 l' LI Ni 0.049 Relative Reduction Potential Reduction Half-equation (oxidized form) ne- (reduced form) 8. Strongest oxidizing agent 9. Strongest reducing agent

Explanation / Answer

Relative reduction potential is the potential for an electrode measured by connecting it to a reference electrode.

given-

reduction potential for your most active metal is 0.0V

Zn2+ +2e --->Zn ,Eo(Zn2+/Zn) anode

Cu2+ +2e--->Cu, Eo(Cu2+/Cu) cathode

Fe2+ +2e--->Fe Eo(Fe2+/Fe)

Ni2+ +2e--->Ni Eo(Ni2+/Ni)

cell Zn+Cu Eo(cell)=0.984V=Eo cathode-Eonode=Eo(Cu2+/Cu)-Eo(Zn2+/Zn)

0.984V=Eo(Cu2+/Cu)-Eo(Zn2+/Zn) -----------------(1)

cell Zn+Fe

Eo(cell)=Eo cathode-Eonode=Eo(Fe2+/Fe)-Eo(Zn2+/Zn) =0.574V

Eo(cell)=Eo(Fe2+/Fe)-Eo(Zn2+/Zn) =0.574V----------(2)

cell Cu+Fe

Eo(cell)=0.497V=Eo cathode-Eonode=Eo(Cu2+/Cu)-Eo(Fe2+/Fe)

or,0.497V=Eo(Cu2+/Cu)-Eo(Fe2+/Fe) .........................(3)

cell Zn+Ni

1.33 V=Eo cathode-Eonode=Eo(Ni2+/Ni)-Eo(Zn2+/Zn) -----------------(4)

cell Cu+Ni

0.304V=Eo(Cu2+/Cu)-Eo(Ni2+/Ni)- -----------------(5)

the eqns (1),2,3,4,5 shows the relative reduction potential

So ,arranging according to relative reduction potential,

Cu Fe Ni Zn

Strongest oxidising agent Cu2+ ,most positive reduction potential

strongest reducing agent Zn ,most negative reduction potential

if Eo(red) for Zn is 0.0V (most active metal ) then the relative red potential are-

then, eqn (1),

0.984V=Eo(Cu2+/Cu)

eqn(2)

Eo(Fe2+/Fe)=0.574V

eqn(4)

1.33 V=Eo(Ni2+/Ni)

7)

Zn2+ +2e --->Zn ,Eo(Zn2+/Zn) =0.0V

Cu2+ +2e--->Cu, Eo(Cu2+/Cu) =0.984V

Fe2+ +2e--->Fe Eo(Fe2+/Fe)=0.574V

Ni2+ +2e--->Ni Eo(Ni2+/Ni)=1.33V

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