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Question
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.
Solution
Given: `"E"_"Mg"^circ` = – 2.37 V and `"E"_"Ag"^circ` = 0.8 V
To find: Standard cell potential
Formula: `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
Calculation: `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
`"E"_"cell"^circ = "E"_"Ag"^circ - "E"_"Mg"^circ`
= (0.8 V) - (- 2.37 V) = 3.17 V
The standard cell potential for the reaction is 3.17 V.
Cell reactions:
Electrode reactions are
| At anode: | \[\ce{Mg_{(s)} -> Mg^{2+}_{ (aq)} + 2e-}\] |
| At cathode: | \[\ce{Ag^+_{ (aq)} + e^- -> Ag_{(s)}}\] |
| Overall cell reaction- \[\ce{Mg_{(s)} + 2Ag^+_{ (aq)} -> Mg^{2+}_{ (aq)} + 2Ag_{(s)}}\] | |
Since the standard cell potential is positive, the cell reaction is spontaneous.
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