A half-cell that consists of a copper wire in 1.00 M CuNO3 solution is connected by a salt bridge to a half cell containing a 1.00 M X(II) acetate solution and an electrode of metal X. The voltage of the cell is 2.00 V, with metal X as the positive electrode. Calculate the standard reduction potential for the X half-reaction. The standard reduction potential of the copper half-reaction is 0.52 V. (Hint, write the half-reactions and the overall chemical equation for the spontaneous reaction.)
To solve this problem, we need to use the Nernst equation and the concept of standard reduction potentials.
The Nernst equation relates the cell potential to the standard reduction potentials and the concentrations of the species involved in the cell reaction:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred in the cell reaction
Q is the reaction quotient (ratio of the concentrations of the products to the concentrations of the reactants)
First, let's write the half-reactions and the overall chemical equation for the spontaneous reaction.
The half-reaction for the copper half-cell is:
Cu2+ + 2e- → Cu
The half-reaction for the X half-cell is:
X(II) + 2e- → X
The overall reaction is obtained by summing these half-reactions:
Cu2+ + X(II) → Cu + X
Now, we have the standard reduction potentials for the copper half-reaction (0.52 V) and want to find the standard reduction potential for the X half-reaction.
Using the given information, we have:
Ecell = 2.00 V (voltage of the cell)
E°Cu = 0.52 V (standard reduction potential of the copper half-reaction)
E°X = ? (standard reduction potential of the X half-reaction)
Since the X(II) acetate solution is the positive electrode, the E°X must be positive.
Using the Nernst equation, we can rearrange it to solve for the standard reduction potential of the X half-reaction:
E°X = Ecell - E°Cu + (0.0592/n) * log(Q)
Since the concentrations are not given, we can assume they are 1.00 M.
Plugging in the values:
E°X = 2.00 V - 0.52 V + (0.0592/n) * log(1.00/1.00)
Simplifying the equation:
E°X = 1.48 V + (0.0592/n) * 0
Since log(1.00/1.00) is zero, we can simplify it further:
E°X = 1.48 V
Therefore, the standard reduction potential for the X half-reaction is 1.48 V.