Write half equation s for the following reaction.
Cu(s)+2AgNO3(aq) produces Cu(NO3)2(aq)+2Ag.
half equations are
Cu ==> Cu^2+ + 2e
Ag^+ e ==>Ag
--------------------------
then
Cu ==> Cu^2+ + 2e
2(Ag^+ e ==>Ag)
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Cu + 2Ag^+ + 2e ==> Cu^2+ + 2e + 2Ag
then cancel 2e and add NO3^-
Cu + 2Ag^+ + 2e ==> Cu^2+ + 2e + 2Ag
NO3^- ==> NO3^-
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Cu + 2AgNO3 ==> Cu(NO3)2 + 2Ag
What would be the standard cell notation for this cell?
Sure, here are the half equations for the given reaction:
Oxidation half equation:
Cu(s) → Cu^2+(aq) + 2e^-
Reduction half equation:
2Ag^+(aq) + 2e^- → 2Ag(s)
Remember, when the oxidation and reduction half equations are combined, we get the overall balanced equation for the reaction!
To write the half equations for the given reaction, we need to separate the oxidation and reduction half-reactions.
Oxidation Half-Reaction:
Cu(s) → Cu2+(aq) + 2e-
Reduction Half-Reaction:
2Ag+(aq) + 2e- → 2Ag(s)
Note: In the oxidation half-reaction, copper (Cu) loses two electrons (2e-) to form a copper ion (Cu2+). In the reduction half-reaction, silver ions (Ag+) gain two electrons (2e-) to form solid silver (Ag).
To write a half equation for the given reaction, we first need to identify which species undergoes oxidation and which undergoes reduction.
In the reaction, copper (Cu) is oxidized as it loses electrons, and silver (Ag) is reduced as it gains electrons.
The half equation for oxidation can be written as follows:
Cu(s) → Cu^2+(aq) + 2e^-
This equation represents the oxidation of copper metal (Cu) to copper(II) ions (Cu^2+) through the loss of two electrons (2e^-).
The half equation for reduction can be written as follows:
2Ag^+(aq) + 2e^- → 2Ag
This equation represents the reduction of silver ions (Ag^+) to silver metal (Ag) through the gain of two electrons (2e^-).
In summary, the half equations for the reaction are:
Oxidation (loss of electrons):
Cu(s) → Cu^2+(aq) + 2e^-
Reduction (gain of electrons):
2Ag^+(aq) + 2e^- → 2Ag