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Question
Answer the following in brief.
Construct a galvanic cell from the electrodes Co3+ | Co and Mn2+/Mn. \[\ce{E{^{\circ}_{Co}}}\] = 1.82 V, \[\ce{E{^{\circ}_{Mn}}}\] = –1.18 V. Calculate \[\ce{E{^{\circ}_{cell}}}\]
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Solution
Given:
\[\ce{E{^{\circ}_{Co}}}\] = 1.82 V,
\[\ce{E{^{\circ}_{Mn}}}\] = –1.18 V.
To find: \[\ce{E{^{\circ}_{cell}}}\] and cell representation
Formulae: `"E"_"cell"^circ = "E"_"Cathode"^circ - "E"_"anode"^circ`
Calculation: Electrode reactions are
At anode: \[\ce{3(Mn_{(s)} -> Mn{^{2+}_{(aq)}} + 2e-}\]
At cathode: \[\ce{2(Co{^{3+}_{(aq)}} + 3e- -> Co_{(s)})}\]
The cell is composed of Mn (anode), Mn(s) \[\ce{Mn{^{2+}_{(aq)}}}\] and Co (cathode), \[\ce{Co{^{3+}_{(aq)}} | Co_{(s)}}\]
The cell is represented as:
\[\ce{Mn_{(s)} | Mn{^{2+}_{(aq)}} || Co{^{3+}_{(aq)}} | Co_{(s)}}\]
The standard electrode potential is given by
\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
= 1.82 V – (–1.18 V)
= 1.82 V + 1.18 V
= 3.00 V
The standard cell potential is 3.00 V.
RELATED QUESTIONS
Choose the most correct option.
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Answer the following in one or two sentences.
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Answer the following:
Predict whether the following reaction would occur spontaneously under standard state condition.
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Calculate `"E"_"cell"^circ` of the following galvanic cell:
Mg(s) / Mg2+(1 M) // Ag+ (1 M) / Ag(s) if `"E"_"Mg"^circ` = – 2.37 V and `"E"_"Ag"^circ` = 0.8 V. Write cell reactions involved in the above cell. Also mention if cell reaction is spontaneous or not.
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Calculate \[\ce{E^0_{cell}}\] for the following cell.
\[\ce{Cr_{(s)} | Cr^{3+}_{( aq)} || Fe^{2+}){( aq)} | Fe_{(s)}}\]
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`"E"_("Fe"^(2+)//"Fe")^0` = −0.44 V
Identify the strongest reducing agent from the data given below:
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| Al | −1.66 |
| Fe | −0.44 |
| Hg | +0.79 |
| Cu | +0.337 |
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\[\ce{Ni_{(s)} | Ni^{2+}_{( aq)} || Au^{3+}_{( aq)} | Au_{(s)}}\]
if \[\ce{E^0_{Ni}}\] = −0.25 V, \[\ce{E^0_{Au}}\] = 1.50 V.
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