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Given the cell reaction:

Ca_((s))+Mg^(2+)_((aq))⇆Ca^(2+)_((aq))+Mg_((s))
Which substance is oxidized?
(1) 
Ca_((s))
(2) 
Mg^(2+)_((aq))
(3) 
Ca^(2+)_((aq))
(4) 
Mg_((s))
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Given the cell reaction:\newlineCa(s)+Mg2+(aq)Ca2+(aq)+Mg(s) \mathrm{Ca}_{(\mathrm{s})}+\mathrm{Mg}^{2+}{ }_{(\mathrm{aq})} \leftrightarrows \mathrm{Ca}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Mg}_{(\mathrm{s})} \newlineWhich substance is oxidized?\newline(11) Ca(s) \mathrm{Ca}_{(s)} \newline(22) Mg2+(aq) \mathrm{Mg}^{2+}{ }_{(a q)} \newline(33) Ca2+(aq) \mathrm{Ca}^{2+}{ }_{(a q)} \newline(44) Mg(s) \mathrm{Mg}_{(\mathrm{s})} \newline11\newline22\newline33\newline44

Full solution

Q. Given the cell reaction:\newlineCa(s)+Mg2+(aq)Ca2+(aq)+Mg(s) \mathrm{Ca}_{(\mathrm{s})}+\mathrm{Mg}^{2+}{ }_{(\mathrm{aq})} \leftrightarrows \mathrm{Ca}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Mg}_{(\mathrm{s})} \newlineWhich substance is oxidized?\newline(11) Ca(s) \mathrm{Ca}_{(s)} \newline(22) Mg2+(aq) \mathrm{Mg}^{2+}{ }_{(a q)} \newline(33) Ca2+(aq) \mathrm{Ca}^{2+}{ }_{(a q)} \newline(44) Mg(s) \mathrm{Mg}_{(\mathrm{s})} \newline11\newline22\newline33\newline44
  1. Identify Oxidation Process: Oxidation is the loss of electrons. Look at the reaction and see which substance loses electrons.\newlineCa(s)Ca(aq)2++2e\text{Ca}_{(s)} \rightarrow \text{Ca}^{2+}_{(aq)} + 2e^-\newlineMg(aq)2++2eMg(s)\text{Mg}^{2+}_{(aq)} + 2e^- \rightarrow \text{Mg}_{(s)}
  2. Determine Electron Loss: Since calcium goes from a neutral state to a 2+2+ charge, it loses 22 electrons. This means calcium is oxidized.

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