An electrochemical cell that uses the reaction

Ti^3+ + 2Cr^2+ -> Ti^1+ + 2Cr^3+

Has a standard cell potential of
E¢ªcell = +1.19V

a. Write the two half-cell reactions.

b. Determine the E¢ªred(reduction).

Anna, I don't understand the symbolism you have used. Perhaps we could help if you corrected that OR if you told us what you don't understand about the problem.

The half cell equations for the cell AS WRITTEN are
Ti^3+ + 2e ==> Ti^+
Cr^2+ ==> Cr^3+ + e

for part b, if the cell as written is 1.19 volts and you want the cell in the opposite direction, that will be -1.19 volts.

To answer part (a) of the question, we need to identify the two half-cell reactions. In an electrochemical cell, there is an oxidation half-reaction and a reduction half-reaction.

In this case, the oxidation half-reaction involves the species Ti^3+, which is being oxidized to Ti^1+. The electron transfer occurs on the left side of the reaction equation, indicating that oxidation is taking place.

The reduction half-reaction involves the species Cr^2+, which is being reduced to Cr^3+. The electron transfer occurs on the right side of the reaction equation, indicating that reduction is taking place.

So, the two half-cell reactions are:

Oxidation half-reaction: Ti^3+ → Ti^1+ + e^-

Reduction half-reaction: 2Cr^2+ + 2e^- → 2Cr^3+

Now, moving on to part (b) of the question, we need to determine the standard reduction potential (E°red) for the reduction half-reaction. The standard reduction potential represents the tendency of a species to be reduced (gain electrons) under standard conditions.

The reduction half-reaction is: 2Cr^2+ + 2e^- → 2Cr^3+

We can look up the standard reduction potential for Cr^3+/Cr^2+ in a table of standard reduction potentials. The standard reduction potential for Cr^3+/Cr^2+ is +0.41V.

Since the standard cell potential (E°cell) is the difference between the standard reduction potentials of the two half-reactions, we can use the equation:

E°cell = E°red (reduction) - E°red (oxidation)

Given that the standard cell potential (E°cell) is +1.19V, and the reduction half-reaction involves Cr^2+/Cr^3+ with an E°red of +0.41V, we can rearrange the equation to solve for E°red (reduction):

E°red (reduction) = E°cell + E°red (oxidation)

E°red (reduction) = +1.19V + (+0.41V)

E°red (reduction) = +1.60V

Therefore, the E°red (reduction) for the given electrochemical cell is +1.60V.