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Question
Answer the following in brief.
Construct a galvanic cell from the electrodes Co3+ | Co and Mn2+/Mn. `"E"_"Co"^circ` = 1.82 V, `"E"_"Mn"^circ` = –1.18 V. Calculate `"E"_"cell"^circ`
Solution
Given:
`"E"_"Co"^circ` = 1.82 V,
`"E"_"Mn"^circ` = –1.18 V.
To find: `"E"_"cell"^circ` and cell representation
Formulae: `"E"_"cell"^circ = "E"_"Cathode"^circ - "E"_"anode"^circ`
Calculation: Electrode reactions are
At anode: `3("Mn"_(("s")) -> "Mn"_(("aq"))^(2+) + 2"e"^-)`
At cathode: `2("Co"_(("aq"))^(3+) + 3"e"^(-) -> "Co"_(("s")))`
The cell is composed of Mn (anode), Mn(s) `|"Mn"_(("aq"))^(2+) and "Co" ("cathode"), "Co"_(("aq"))^(3+)| "Co"_(("s"))`
The cell is represented as:
`"Mn"_(("s")) |"Mn"_(("aq"))^(2+)| |"Co"_(("aq"))^(3+)| "Co"_(("s"))`
The standard electrode potential is given by
`"E"_"cell"^circ = "E"_"Cathode"^circ - "E"_"anode"^circ`
= 1.82 V – (–1.18 V)
= 3.00 V
The standard cell potential is 3.00 V.
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