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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.
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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.
RELATED QUESTIONS
Choose the most correct option.
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Given: `"E"_("Cr"^(3+)//"Cr")^0` = −0.74 V,
`"E"_("Fe"^(2+)//"Fe")^0` = −0.44 V
Identify the strongest reducing agent from the data given below:
| Element | `"E"^0 ("V")` |
| 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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If \[\mathrm{E^o\left(Cd_{(aq)}^{+2}|Cd_{(s)}\right)=-0.40~V}\]. What is potential for \[\mathrm{Cd}_{(s)}\xrightarrow{}\mathrm{Cd}_{(aq)}^{+2}\left(0.01\mathrm{M}\right)+2\mathrm{e}^{-}\] at 298 K?
The cell potential for the following cell notation is approximately:
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