A piece of cadmium is placed in a solution in which [Cd^2+]=[Sn^2+]=1.00M. Will the following reaction proceed spontaneously?
Cd(s)+Sn^2+(aq)-->Cd^2+(aq)+Sn(s)
Look up the Eo values, add Eo oxidation + Eo reduction. If the number is + the reaction will go; if negative, the reaction will not go as written.
Well, if cadmium and tin were to get married, would they have a "snappy" wedding? I'm not sure, but let's take a closer look at the chemistry behind it.
In this reaction, cadmium is being oxidized from its elemental state (Cd(s)) to Cd^2+(aq), while tin is being reduced from Sn^2+(aq) to its elemental state (Sn(s)).
To determine if the reaction will proceed spontaneously, we need to examine the standard reduction potentials of both reactions. If the reduction potential of the cadmium half-reaction is greater than that of the tin half-reaction, then the reaction should proceed spontaneously.
Unfortunately, I seem to have misplaced my periodic table, so I can't check the values. But hey, maybe it's for the best. I wouldn't want to cause a scientific catastrophe by mixing up some elements!
In any case, I can't give you a definitive answer, but I hope I was able to bring a smile to your face with my silly antics!
To determine if the reaction will proceed spontaneously, we can compare the standard reduction potentials of the two half-reactions involved:
Half-reaction 1: Cd^2+(aq) + 2e^- → Cd(s)
Half-reaction 2: Sn^2+(aq) + 2e^- → Sn(s)
The spontaneous reaction occurs when the reduction potential of the half-reaction with the more positive reduction potential is greater than the reduction potential of the half-reaction with the less positive reduction potential.
Looking up the standard reduction potentials for Cd^2+ and Sn^2+ in a table, we find that:
E°(Cd^2+/Cd) = -0.40 V
E°(Sn^2+/Sn) = -0.14 V
Since -0.14 V is larger (less negative) than -0.40 V, the reduction potential for Sn^2+/Sn is more positive than Cd^2+/Cd. Therefore, reaction Cd(s) + Sn^2+(aq) → Cd^2+(aq) + Sn(s) will not proceed spontaneously.
To determine if the reaction will proceed spontaneously, we need to compare the standard reduction potentials of the half-reactions involved. The reaction will proceed spontaneously if the overall cell potential (Ecell) is positive.
The half-reactions for the species involved in the reaction are as follows:
Cd2+(aq) + 2 e- → Cd(s) (1)
Sn2+(aq) + 2 e- → Sn(s) (2)
The standard reduction potentials for these half-reactions can be found in tables of reduction potentials. Let's assume that the values are:
E°(Cd2+/Cd) = -0.40 V
E°(Sn2+/Sn) = -0.14 V
To determine the overall cell potential (Ecell) of the reaction, we need to subtract the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs):
Ecell = E°(cathode) - E°(anode)
In this case, Cd is being reduced and Sn is being oxidized. Therefore, Cd2+/Cd is the cathode half-reaction and Sn2+/Sn is the anode half-reaction.
Ecell = E°(Cd2+/Cd) - E°(Sn2+/Sn)
Ecell = (-0.40 V) - (-0.14 V)
Ecell = -0.40 V + 0.14 V
Ecell = -0.26 V
Since Ecell is negative (-0.26 V), the reaction will not proceed spontaneously in the given conditions.