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Question
Calculate standard cell potential of following galvanic cell:
Zn/Zn2+(1 M) // Pb2+(1 M)/Pb. If `"E"_"Pb"^0` = 0.126 V and `"E"_"Zn"^0` = –0.763 V
Solution
Given: `"E"_"Pb"^0` = 0.126 V, `"E"_"Zn"^0` = - 0.763 V
To find: Standard cell potential
Formula: `"E"_"cell"^0 = "E"_"cathode"^0 - "E"_"anode"^0`
Calculation: Electrode reactions are
At anode: \[\ce{Zn_{(s)} -> Zn^{2+}_{ (aq)} + 2e-}\]
At cathode: \[\ce{Pb^{2+}_{ (aq)} + 2e- -> Pb_{ (s)}}\]
The standard cell potential is given by
`"E"_"cell"^0 = "E"_"cathode"^0 - "E"_"anode"^0`
`"E"_"cell"^0 = "E"_"Pb"^0 - "E"_"Zn"^0` = (0.126 V) – –0.763 V) = 0.889 V
The standard cell potential for the reaction is 0.889 V.
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