Using standard reduction potentials, calculate the standard E¡Æcell, emf, for each of the following reactions. State if each reaction is spontaneous or nonspontaneous. (Use the below reactions AS WRITTEN, DO NOT REARRANGE ANY REACTIONS.) You must show the two half-reactions and balance the gain and loss of electrons.

a. H2(g) + I2(s) -> 2H^1+(aq) + 2I^1-(aq)

b. Ni^2+(aq) + 2Ce^5+(aq) -> Ni¡Æ(s) + 2Ce^4+(aq)

c. 4Ce^3+(aq) -> Ce¡Æ(s) + 3Ce^4+(aq) (Acid solution)

d. 2Al^3+(aq) + 3Cd(s) -> 2Al(s) + 3Cd^2+(aq)

To calculate the standard cell potential (E°cell) and determine if a reaction is spontaneous or nonspontaneous, we need to follow a few steps:

1. Identify the half-reactions: Break down the overall reaction into two half-reactions, one representing the reduction and the other representing oxidation. This step involves assigning oxidation numbers and balancing the electrons gained and lost in each half-reaction.

2. Lookup reduction potentials: Use a table of standard reduction potentials to find the reduction potential (E°red) value for each half-reaction. The reduction potential represents the tendency of a reduction half-reaction to occur. The more positive the reduction potential, the stronger the tendency.

3. Calculate overall E°cell: Subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction to obtain the overall E°cell value. The sign of E°cell indicates the spontaneity of the reaction. If E°cell is positive, the reaction is spontaneous, while a negative E°cell indicates a nonspontaneous reaction.

Let's go through the process for each reaction:

a. H2(g) + I2(s) -> 2H^1+(aq) + 2I^1-(aq)

Oxidation half-reaction: H2(g) -> 2H^1+(aq) + 2e^-
Reduction half-reaction: I2(s) + 2e^- -> 2I^1-(aq)

Lookup the reduction potentials in a standard reduction potential table:
E°red(H+ / H2) = 0.00 V
E°red(I- / I2) = 0.54 V

Overall E°cell = E°red(Reduction) - E°red(Oxidation)
= 0.54 V - 0.00 V
= 0.54 V

Since E°cell is positive, the reaction is spontaneous.

b. Ni^2+(aq) + 2Ce^5+(aq) -> Ni¡Æ(s) + 2Ce^4+(aq)

Oxidation half-reaction: Ce^5+(aq) -> Ce^4+(aq) + e^-
Reduction half-reaction: Ni^2+(aq) + 2e^- -> Ni¡Æ(s)

Lookup the reduction potentials:
E°red(Ce^5+ / Ce^4+) = 1.61 V
E°red(Ni^2+ / Ni) = -0.25 V

Overall E°cell = E°red(Reduction) - E°red(Oxidation)
= (-0.25 V) - (1.61 V)
= -1.86 V

Since E°cell is negative, the reaction is nonspontaneous.

c. 4Ce^3+(aq) -> Ce¡Æ(s) + 3Ce^4+(aq) (Acid solution)

Oxidation half-reaction: Ce^3+(aq) -> Ce^4+(aq) + e^-
Reduction half-reaction: Ce^4+(aq) + e^- -> Ce¡Æ(s)

Lookup the reduction potentials:
E°red(Ce^4+ / Ce^3+) = 1.61 V

Overall E°cell = E°red(Reduction) - E°red(Oxidation)
= 0 V - 1.61 V
= -1.61 V

Since E°cell is negative, the reaction is nonspontaneous.

d. 2Al^3+(aq) + 3Cd(s) -> 2Al(s) + 3Cd^2+(aq)

Oxidation half-reaction: Cd(s) -> Cd^2+(aq) + 2e^-
Reduction half-reaction: Al^3+(aq) + 3e^- -> Al(s)

Lookup the reduction potentials:
E°red(Cd^2+ / Cd) = -0.40 V
E°red(Al^3+ / Al) = -1.66 V

Overall E°cell = E°red(Reduction) - E°red(Oxidation)
= (-1.66 V) - (-0.40 V)
= -1.26 V

Since E°cell is negative, the reaction is nonspontaneous.

Remember that the spontaneity of a reaction can also be determined by the sign of ΔG°, the standard Gibbs free energy change. A negative ΔG° indicates a spontaneous reaction, while a positive ΔG° indicates a nonspontaneous reaction.

a. The two half-reactions for the first reaction are:

1. Oxidation half-reaction: H2(g) -> 2H^1+(aq) + 2e^-
2. Reduction half-reaction: I2(s) + 2e^- -> 2I^1-(aq)

Now, let's check the reduction potentials for each half-reaction:

Reduction potential for half-reaction 1: H2(g) -> 2H^1+(aq) + 2e^- = 0.00 V
Reduction potential for half-reaction 2: I2(s) + 2e^- -> 2I^1-(aq) = 0.54 V

To calculate the standard cell potential (E°cell), use the formula:
E°cell = E°reduction (cathode) - E°reduction (anode)

E°cell = 0.54 V - 0.00 V = 0.54 V

Since the standard cell potential is positive (0.54 V), the reaction is spontaneous.

b. The two half-reactions for the second reaction are:

1. Oxidation half-reaction: Ni^2+(aq) -> Ni(s) + 2e^-
2. Reduction half-reaction: Ce^5+(aq) + e^- -> Ce^4+(aq)

Now, let's check the reduction potentials for each half-reaction:

Reduction potential for half-reaction 1: Ni^2+(aq) -> Ni(s) + 2e^- = -0.26 V
Reduction potential for half-reaction 2: Ce^5+(aq) + e^- -> Ce^4+(aq) = 1.61 V

To calculate the standard cell potential (E°cell), use the formula:
E°cell = E°reduction (cathode) - E°reduction (anode)

E°cell = 1.61 V - (-0.26 V) = 1.87 V

Since the standard cell potential is positive (1.87 V), the reaction is spontaneous.

c. The two half-reactions for the third reaction are:

1. Oxidation half-reaction: 4Ce^3+(aq) -> Ce(s) + 3Ce^4+(aq) + 3e^-

Now, let's check the reduction potential for the half-reaction:

Reduction potential for the half-reaction: Ce^4+(aq) + e^- -> Ce^3+(aq) = 1.61 V

Since there is only one half-reaction given, we cannot calculate the standard cell potential (E°cell). However, we can determine if the reaction is spontaneous or nonspontaneous by comparing the reduction potential to zero. In this case, the reduction potential (1.61 V) is positive, so the reaction is spontaneous.

d. The two half-reactions for the fourth reaction are:

1. Oxidation half-reaction: 3Cd(s) -> 3Cd^2+(aq) + 6e^-
2. Reduction half-reaction: 2Al^3+(aq) + 6e^- -> 2Al(s)

Now, let's check the reduction potentials for each half-reaction:

Reduction potential for half-reaction 1: Cd^2+(aq) + 2e^- -> Cd(s) = -0.40 V
Reduction potential for half-reaction 2: Al^3+(aq) + 3e^- -> Al(s) = -1.66 V

To calculate the standard cell potential (E°cell), use the formula:
E°cell = E°reduction (cathode) - E°reduction (anode)

E°cell = -1.66 V - (-0.40 V) = -1.26 V

Since the standard cell potential is negative (-1.26 V), the reaction is nonspontaneous.