why is this reaction spontaneous?
Mg + 2Cu2+ --> Mg2+ + 2Cu+
Mg2+ + 2e --> Mg E= -2.37 V
Cu2 + e --> Cu+ E= 0.15 V
Hi Tara,
To find out if a reaction is spontaneous, we calculate E(cell). If E(cell)>0, reaction is spontaneous. If E(cell)<0, reaction is not spontaneous. If E(cell)=0, reaction will proceed until equilibrium is established.
So in this case, E(cell) = E(reduction)-E(oxidation) = 0.15V - (-2.37V) = 2.52V >0 hence reaction is spontaneous.
In this case, Mg is oxidized to Mg2+ as it loses 2 electrons while Cu2+ is reduced to Cu+ as it gains 1 electron.
Hope I helped! (:
-J
To determine whether a reaction is spontaneous or not, we can look at the standard cell potential. The standard cell potential (Ecell) is the measure of the driving force of a reaction and determines whether the reaction is spontaneous or non-spontaneous.
In this case, we have two half-reactions:
1. Mg2+ + 2e --> Mg (E = -2.37 V)
2. Cu2+ + e --> Cu+ (E = 0.15 V)
To calculate the standard cell potential, we need to assign the appropriate signs to the reduction potentials. In this case, the Cu2+ + e --> Cu+ reaction is a reduction, and the Mg2+ + 2e --> Mg reaction is oxidation, so the signs are already assigned correctly.
The standard cell potential (Ecell) is calculated by subtracting the reduction potential of the anode (oxidation) from the reduction potential of the cathode (reduction).
Ecell = E(cathode) - E(anode)
Ecell = 0.15 V - (-2.37 V)
Ecell = 0.15 V + 2.37 V
Ecell = 2.52 V
Since the Ecell value is positive (greater than zero), it indicates that the reaction is spontaneous. A positive value for Ecell means that the reaction will proceed forward spontaneously, and if the value is negative, it means that the reaction is non-spontaneous and will require an external driving force to occur.
Therefore, based on the positive Ecell value of 2.52 V, we can conclude that the given reaction (Mg + 2Cu2+ --> Mg2+ + 2Cu+) is spontaneous.