Given the following two half-reactions, determine which overall reaction is spontaneous and calculate its standard cell potential.
Cd^2+(aq) + 2e^- ==> Cd(s) E naught = -1.21V
Sn^2+ + 2e^- ==Sn(s) E naught = -0.14V
Sn^2+ +Cd(s) ==> Sn(s) + Cd^2+
delta e prime-naught will be 1.07V
Do you mean you want to determine which way the cell is spontaneous?
Add an oxidation half to a reduction half, the one that ends up with a positive Ecell is spontaneous. If I have misinterpreted the question, please clarify.
To determine which overall reaction is spontaneous, we can use the concept of standard cell potentials. The standard cell potential, denoted as E°, is the difference in reduction potentials between the two half-reactions.
The reaction is spontaneous if the overall standard cell potential, E°, is positive.
In this case, we have:
Cd^2+(aq) + 2e^- ==> Cd(s) E° = -1.21V
Sn^2+ + 2e^- ==> Sn(s) E° = -0.14V
To calculate the overall standard cell potential, we need to determine the reduction potential of each half-reaction.
Since the reduction potential of the Cd^2+/Cd half-reaction (-1.21V) is more negative than the reduction potential of the Sn^2+/Sn half-reaction (-0.14V), the Cd^2+/Cd half-reaction is the reduction half-reaction, and the Sn^2+/Sn half-reaction is the oxidation half-reaction.
To calculate the overall standard cell potential, subtract the oxidation potential from the reduction potential:
E°overall = E°reduction - E°oxidation
E°overall = (-1.21V) - (-0.14V)
E°overall = -1.21V + 0.14V
E°overall = -1.07V
Since the overall standard cell potential is negative (-1.07V), the overall reaction is not spontaneous.
To determine which overall reaction is spontaneous, we can compare the standard cell potentials (E°) of the two half-reactions. The reaction with the more positive E° value will be the spontaneous reaction.
In this case, we have two half-reactions:
1) Cd^2+(aq) + 2e^- ==> Cd(s) with E° = -1.21V
2) Sn^2+ + 2e^- ==> Sn(s) with E° = -0.14V
To calculate the standard cell potential (E°cell) for the overall reaction, we need to subtract the E° value of the second half-reaction from the E° value of the first half-reaction.
E°cell = E° (reduction) - E° (oxidation)
In this case, the first half-reaction is the reduction of Cd^2+(aq) and the second half-reaction is the reduction of Sn^2+.
E°cell = E° (reduction of Cd^2+(aq)) - E° (reduction of Sn^2+)
= (-1.21V) - (-0.14V)
= -1.07V
The negative value of the standard cell potential (-1.07V) indicates that the overall reaction is spontaneous in the opposite direction of the written reaction.
Therefore, the overall reaction is:
2Cd(s) + Sn^2+(aq) => 2Cd^2+(aq) + Sn(s)
with a standard cell potential of -1.07V.